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Edit 2: After thinking through this again, I realized that my argument for the final case, when $\tau$ has both a left and right endpoint, contained an error. I've given a new, simpler argument for that case, using Lindenbaum's version of the Schroeder-Bernstein theorem for linear orders. The argument applies to orders of arbitrary size: it shows that if $\tau$ is any order (not necessarily countable) with both a left and right endpoint, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$.

The two-endpoint case seems special: the argument does not seem to apply to uncountable $\tau$ with no endpoints, or with a single endpoint. So those cases remain open.

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Edit: As Joel pointed out in the comments, in the previous version of my answer, I reversed the usual meaning of $\times$ for linear orders. I've now hopefully corrected all instances of the reversal.

He's also commented that lifting this proof from countable $\tau$ to general $\tau$ may be impossible, so for now this is only a partial answer.

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For countable $\tau$, I believe the answer is no. Joel Hamkins has commented on your Math.SE version of this question that from this we can get a negative answer for general $\tau$ by a forcing argument, giving a complete answer to your question. Joel, if you read this, it'd be great if you could give an explanation of your comment. (Update: see Joel's comments and the edit above.)

Theorem. If $\tau$ is a countable linear order, and for some $n$ we have $\tau^n \cong \tau$, then $\tau^2 \cong \tau$.

Proof. There are four possibilities: either $\tau$ has no endpoints, $\tau$ has a left endpoint but not a right one, $\tau$ has a right endpoint but not a left one, or $\tau$ has both a left and right endpoint.

Case 1: Assume first that $\tau$ has no endpoints. We'll handle the other cases in turn.

The idea is to use the hypothesis $\tau^n \cong \tau$ to decompose $\tau$ in a certain way, and then use this decomposition to show $\tau^2 \cong \tau$. Forgetting $\tau$ for a moment, let's describe the form of the decomposition we are aiming for. Let $\eta$ denote the set of rationals. Partition $\eta$ into countably many disjoint dense sets, that is, find subsets $\eta_i \subseteq \eta$ such that each $\eta_i$ is dense, $\eta = \bigcup_{i \in \mathbb{N}} \eta_i$, and for $i \neq j$ we have $\eta_i \cap \eta_j = \emptyset$. I'll refer to the elements of $\eta_i$ as $i$-points. We think of $i$-points as being colored by the color $i$.

Now, suppose for each $i \in \mathbb{N}$ we are given some countable linear order $L_i$. We may form a new order $L$ by replacing, for each $i \in \mathbb{N}$, each $i$-point in $\eta$ with a copy of $L_i$. Somewhat informally, I'll write \begin{equation} L = \bigcup_{i \in \mathbb{N}} L_i \times \eta_i. \end{equation} Notice that the three examples you gave of infinite orders satisfying $\tau^2 \cong \tau$ are all of the form of $L$. To get $\eta$ for example, let $L_i = 1$ for every $i$. The orders $\omega \times \eta$ and $\omega^2 \times \eta$ are respectively obtained by letting $L_i = \omega$ for every $i$, and $L_i = \omega^2$ for every $i$. A more complicated order of this form is obtained by letting $L_0 = \omega$, and $L_i = \omega^2$ for $i > 0$. One may think of this order as $\omega \times \eta$ interspersed with $\omega^2 \times \eta$.

Lemma. For $L$ of this form, we always have $L^2 \cong L$. Indeed, if $\beta$ is any countable order, we have $L \times \beta \cong L$.

Proof. One may think of $L \times \beta$ as being formed in two steps. First, replace each point in $\beta$ with copy of $\eta$ to form $\eta \times \beta$. Since each copy of $\eta$ is partitioned into $i$-points, we may think of $\eta \times \beta$ as also being partitioned into $i$-points. Then for each $i$, the collection of $i$-points is dense in $\eta \times \beta$. Now, since $\beta$ is countable, $\eta \times \beta$ is isomorphic to $\eta$. Moreover, there is an isomorphism between $\eta \times \beta$ and $\eta$ that sends $i$-points to $i$-points. (I am using here a stronger form of Cantor's theorem that all countable dense orders without endpoints are isomorphic to $\eta$. Namely, if $X$ and $Y$ both have order type $\eta$, and each is decomposed into countably many disjoint dense sets, $X = \bigcup_i X_i$, $Y = \bigcup_i Y_i$, then there is an isomorphism of $X$ and $Y$ that sends $X_i$ onto $Y_i$ for every $i$.) Now, $L \times \beta$ is formed by replacing every $i$-point in $\eta \times \beta$ with a copy of $L_i$. Since $L$ is formed by replacing every $i$-point in $\eta$ with a copy of $L_i$, and there is an isomorphism between $\eta$ and $\eta \times \beta$ that respects our coloring, we must have $L \times \beta \cong L$. This proves the lemma. Notice that the same argument works if some (or even all but finitely many) of the $L_i$ are empty.

Our goal, then, is to show that $\tau^n \cong \tau$ implies that $\tau$ may be decomposed into the form of $L$; that is, $\tau = \bigcup_i L_i \times \eta_i$ for some collection of countable order types $L_i$. Then by the claim we will have $\tau^2 \cong \tau$. Let us write our hypothesis as $\tau \cong \tau \times \beta$, where $\beta = \tau^{n-1}$. For us, the form of $\beta$ is irrelevant except that it is countable and, like $\tau$, has neither a left nor right endpoint.

The fact that $\tau \cong \tau \times \beta$, means that $\tau$ may be split up into $\beta$-many intervals, each of order type $\tau$. Each of these $\beta$-many copies of $\tau$ may in turn be split into $\beta$-many copies of $\tau$, and so on. This is as much to say that to every finite sequence $s = \langle x_1, x_2, \ldots, x_n \rangle$ of elements of $\beta$, we may associate an interval $I_s$ that is isomorphic to $\tau$. Namely, $I_s$ is the $x_n$-th copy of $\tau$ within the $x_{n-1}$-th copy of $\tau$ $\ldots$ within the $x_1$-st copy of $\tau$. Since every $I_s$ is of order type $\tau$, we also have some $f_s: \tau \rightarrow I_s$ witnessing the isomorphism.

In iteratively splitting $\tau$ into smaller and smaller copies of itself, we may choose the $I_s$ and $f_s$ in such a way that the maps $f_s$ respect concatenation, that is, so that if $s^{\frown}t$ is the sequence obtained by concatenating the sequences $s$ and $t$, then $f_{s^{\frown}t} = f_{s} \circ f_{t}$ (where, on the right, $f_s$ really means $f_s \upharpoonright I_t$). Here's how to do this. At the first stage, for every $x \in \beta$ we have an interval $I_{\langle x \rangle}$ of order type $\tau$ and an isomorphism $f_{\langle x \rangle}: \tau \rightarrow I_{\langle x \rangle}$. We define $I_s$ and $f_s$ for all longer $s$ in terms of these first-level $f_{\langle x \rangle}$ and $I_{\langle x \rangle}$. At the second stage, for every $y \in \beta$, let $I_{\langle x, y \rangle}$ be the image of $I_{\langle y \rangle}$ under $f_{\langle x \rangle}$. Then we simply define $f_{\langle x, y \rangle}$ to be$ f_{\langle x \rangle} \circ f_{\langle y \rangle}$, which by our choice of $I_{\langle x, y \rangle}$ is an isomorphism $\tau$ onto $I_{\langle x, y \rangle}$. For every $z \in \beta$, let $I_{\langle x, y, z \rangle} = f_{\langle z \rangle}[I_{\langle x, y \rangle}]$, and and $f_{\langle x, y, z \rangle} = f_{\langle x \rangle} \circ f_{\langle y \rangle} \circ f_{\langle z \rangle}$. And so on for longer sequences. The $f_s$ so defined clearly respect concatenation. Note that if $s$ extends $t$, then $I_s$ is a proper subinterval of $I_t$.

It may be helpful to illustrate this construction with an example. If, say, $X = [0, 1)$, then $X \cong X \times 2$, as witnessed by splitting up $X$ as $[0, \frac{1}{2}) \cup [\frac{1}{2}, 1)$. Here, $2=\{0, 1\}$ is playing the role of $\beta$, and we have $I_0 = [0, \frac{1}{2})$, and $I_1=[\frac{1}{2}, 1)$. Let $f_0(x) = \frac{1}{2}x$ be our isomorphism from $X$ onto $I_0$ and $f_1(x) = \frac{1}{2}x + \frac{1}{2}$ be our isomorphism from $X$ onto $I_1$. Then, for example, $I_{01} = f_0[I_1] = [\frac{1}{4}, \frac{1}{2})$ and $f_{01} = f_0 \circ f_1 = \frac{1}{2}(\frac{1}{2} x + \frac{1}{2}) = \frac{1}{4}x + \frac{1}{4}$ is our isomorphism from $X$ onto $I_{01}$. By taking longer compositions of $f_0$ and $f_1$, we obtain $I_s$ and $f_s$ for every $s \in 2^{<\omega}$.

Now, back in our setting, having associated to every finite sequence $s \in \beta^{<\omega}$ an interval $I_s$, we may associate to every infinite sequence $r \in \beta^{\omega}$ the interval $I_r = \bigcap_{n} I_{r \upharpoonright n}$ obtained by taking the natural nested intersection. The collection of $I_r$ for $r \in \beta^{\omega}$ is a covering of $\tau$ by nonintersecting intervals. Let us view $\beta^{\omega}$ also as a linear order, under the lexicographical ordering of sequences. Then our construction guarantees that $I_r$ lies to the left of $I_{r'}$ in $\tau$ if and only if $r <_{lex} r'$ in $\beta^{\omega}$. Indeed, $\tau$ is recovered from $\beta^{\omega}$ by replacing every $r \in \beta^{\omega}$ by the corresponding interval $I_r$. Note, however, that for a fixed $r$, it may be that $I_r$ contains many points, a single point, or no points at all. In fact, since $\tau$ itself is countable, $I_r$ will be empty for all but countably many $r$.

The relevant observation is that, if we view the interval $I_r$ as a linear order, then for densely many $s \in \beta^{\omega}$ we have $I_r \cong I_s$. To see this, let us introduce an equivalence relation on the space $\beta^{\omega}$: say that $r \sim s$ iff $r$ and $s$ share a tail-sequence (not necessarily beginning at the same coordinate). That is: \begin{equation} r \sim s \leftrightarrow \textrm{there exists $m, n$ such that for every $i$ we have $r(m+i) = s(n+i)$}, \end{equation} where $r(k)$ means the $k$th entry in the sequence $r$, etc. This is clearly an equivalence relation. Notice that since $\beta$ is countable, every $\sim$-equivalence class is countable.

Claim. If $r \sim s$, then $I_r \cong I_s$.

Proof. If $r \sim s$, then there are finite sequence $r', s' \in \beta^{<\omega}$ and an infinite sequence $x \in \beta^{\omega}$ such that $r = r'^{\frown}x$ and $s=s'^{\frown}x$. The intuitive reason that $I_r \cong I_s$ is that, in order to construct these intervals, we first move to $I_{r'}$ and $I_{s'}$, which are both just copies of $\tau$. Then, in each interval, we travel down the precisely corresponding nested sequence of intervals (represented by $x$) to obtain $I_r$ and $I_s$, which must therefore be isomorphic. More formally, we see that by the way we defined $f_{r'}$ and $f_{s'}$, we have $f_{s'} \circ f_{r'}^{-1}$ maps $I_r$ isomorphically onto $I_s$, establishing the claim.

For $s \in \beta^{\omega}$, let $\bar{s}$ denote its equivalence class. It is easy to see that if $s_1, s_2 \in \bar{s}$ and $s_1 < s_2$ in the lexicographical order, then there are $r, r', r'' \in \bar{s}$ such that $r < s_1 < r' < s_2 < r''$ (we use here that $\beta$ has neither a top nor bottom element). Since $\bar{s}$ is countable, it follows that $\bar{s}$ (viewed as a suborder of $\beta^{\omega}$) has order type $\eta$. In fact, it's similarly easy to see that $\bar{s}$ is not only dense in its order type, but is actually dense in $\beta^{\omega}$.

Now by the claim, for a fixed $\bar{s}$, every $I_r$ with $r \in \bar{s}$ is isomorphic. Since only countably many of the $I_r$ are nonempty, we may enumerate the (distinct) classes $\overline{s_1}, \overline{s_2}, \ldots$, such that if $I_r$ is nonempty, there is $i$ such that $r \in \overline{s_i}$ (this list may even be finite). Let $L_i$ be the shared order type of every $I_r$, $r \in \overline{s_i}$.

But now we have our desired decomposition. As already noted, every $\overline{s_i}$ has type $\eta$, and is dense in $\beta^{\omega}$. Thus the subset $S \subseteq \beta^{\omega}$ given by $S = \bigcup_i \overline{s_i}$ also has order type $\eta$. We think of each $\overline{s}_i \subseteq S$ as our collection of $i$-points. But then since the $\overline{s_i}$ range over all nonempty intervals $I_r$, we have that $\tau$ may be formed by replacing each $i$-point in $S$ by a copy of $L_i$. That is, $\tau = \bigcup_i \overline{s_i} \times L_i$. This is a decomposition of the sought after form. By our lemma at the beginning, we have $\tau^2 \cong \tau$, as desired.

This takes care of the case when $\tau$ has neither a left nor right endpoint. The case when $\tau$ has a single endpoint is similar, but with complicating details.

Case 2 (sketch): Suppose now that $\tau$ has a left endpoint, but not a right one. The case where $\tau$ has a right endpoint but not a left one is symmetric.

We seek to adapt the proof of case 1. The preliminary discussion preceding the lemma is modified as follows. Instead of decomposing $\eta$ into countably many copies of itself, we decompose $1+\eta$ (that is, the order type of the rationals with a left endpoint added) as $1 + \eta_0 \, \, \cup \, \, \,\bigcup_{i>0} \eta_i$, where the $\eta_i$ are pairwise disjoint, and each is dense in $\eta$. This is the same as before, except that in our decomposition we have the distinguished set $1+\eta_0$ that includes the left endpoint, i.e. we color the left endpoint as a $0$-point. Suppose as before we are given a collection of countable orders $L_i$, where now we insist $L_0$ has a left endpoint. We may form $L$ by replacing every $i$-point with $L_i$, and write \begin{equation} L = L_0 \times (1 + \eta_0) \, \, \cup \, \, \bigcup_{i>0} L_i \times \eta_i. \tag{$\star$} \end{equation}

Note that since $L_0$ has a left endpoint, $L$ does as well. The claim is again that $L^2 \cong L$, or more generally that if $\beta$ is any countable order also with a left endpoint, then $L \times \beta \cong L$. The same argument as before goes through, the relevant difference being that whereas in the previous case we used the fact that $\eta \times \beta \cong \eta$ for any countable $\beta$, we now use the fact that since $\beta$ has a left endpoint, we have $(1+\eta) \times \beta \cong 1+\eta$ (so that $(1+\eta) \times \beta$ may be colored in the same way we colored $1+\eta$).

Then, we seek to use $\tau^n \cong \tau$ to decompose $\tau$ in the form of ($\star$). Again we write $\tau$ as $\tau \times \beta$, where $\beta = \tau^{n-1}$, and note that $\beta$ has a left endpoint, which we label as 0. We again associate to every $s \in \beta^{<\omega}$ an interval $I_s$ and isomorphism $f_s$ of $I_s$ with $\tau$. By taking intersections we obtain the intervals $I_s$ for $s \in \beta^{\omega}$. Again if $s \sim r$, then $I_s \cong I_r$.

For each $s \in \beta^{\omega}$, the equivalence class $\bar{s}$ will again have the order type of the rationals, except for the equivalence class of the zero sequence $\mathbf{0} = \langle 0, 0, \ldots \rangle$. This class has order type $1 + \eta$, the left endpoint being $\mathbf{0}$ itself. Label this class as $\overline{s_0}$. Let $L_0$ be the shared order type of intervals associated to this class, and note that $L_0$ has a left endpoint. For $L_0$ is the order type of $I_{\mathbf{0}}$, which contains the left endpoint of $\tau$. We may enumerate the other classes corresponding to nonempty intervals as $\overline{s_1}, \overline{s_2}, \ldots$. Letting $L_i$ be the order type of the intervals associated to $\overline{s_i}$, we again see that $\tau$ may be written as $(L_0 \times \overline{s_0}) \cup \bigcup_{i>0} L_i \times \overline{s_i}$. Since $\overline{s_0}$ has type $1+\eta$ and all other $\overline{s_i}$ have type $\eta$, this is a decomposition of the desired form ($\star$).

This completes the proof of case 2. As noted, the argument for when $\tau$ has a right endpoint but not a left one is symmetric.

Case 3: Now suppose that $\tau$ has both a left and right endpoint.

Update: I originally gave an analogous argument for this case, but there is a detail that prevents the argument from succeeding. Here is the error. In the first case, when $\tau$ had no endpoints, we used the fact that for an arbitrary countable $\beta$, we have $\eta \times \beta \cong \eta$. In the second, we noted that as long as $\beta$ has a left endpoint, $(1+\eta) \times \beta \cong 1 + \eta$. For this final case, the analogous claim would be that if $\beta$ has both a left and right endpoint, then $(1+\eta+1)\times \beta \cong 1+\eta+1$. While this is true if $\beta$ is itself $1+\eta+1$, it is false in general; indeed it is false for any $\beta$ that is not dense. In particular, while the other details of the argument go through, it does not follow that $\tau^2 \cong \tau$, and I do not see how this argument can be fixed.

New argument: We can still show that $\tau^2 \cong \tau$ in this case, but by an entirely different argument. The lemma we will need is Lindenbaum's version of the Schroeder-Bernstein theorem for linear orders. It says: if $X$ and $Y$ are linear orders such that $X$ is isomorphic to an initial segment of $Y$, and $Y$ is isomorphic to a final segment of $X$, then in fact $X \cong Y$. The proof of the theorem is the same as the usual one; the hypotheses guarantee that the bijection constructed between $X$ and $Y$ is order-preserving.

So suppose that $\tau$ has both a left and right endpoint, and for some $n$, $\tau^n \cong \tau$. We show $\tau^2 \cong \tau$ by Lindenbaum. Since $\tau$ has a left endpoint, we see that, viewing $\tau^2$ as $\tau$-many copies of $\tau$, that $\tau^2$ contains an initial copy of $\tau$. That is, $\tau$ is isomorphic to an initial segment of $\tau^2$. We will be done if we can show that $\tau^2$ is isomorphic to a final segment of $\tau^n$, since $\tau^n \cong \tau$. Viewing $\tau^n$ as $\tau^{n-2}$ copies of $\tau^2$, and using the fact that $\tau^{n-2}$ has a right endpoint (as $\tau$ does), we see that $\tau^n$ has a final copy of $\tau^2$, that is, $\tau^2$ maps isomorphically onto a final segment of $\tau^n$, as desired.

In no part of this argument did we need that $\tau$ was countable, so that indeed if $\tau$ is an order of arbitrary size with both endpoints, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$. Unfortunately, it seems impossible to use a similar argument in the other cases, since to apply Lindenbaum, we really need that $\tau$ has both endpoints.

Update 10/4/16: There is no such order type.

Theorem. Let $X$ be a linear order. If $X^n \cong X$ for some $n>1$, then $X^2 \cong X$.

My previous answer, which dealt with the countable case, can be found in the edit history.

For the proof, see my paper. There's an overview of the first part of the proof in these slides. In both the paper and the slides, I use the lexicographical ordering on the product of two orders.

Very roughly, the argument goes by showing that if $X^n \cong X$ for some $n>2$, then it is possible to construct an isomorphism between $X^2$ and $X$ by stitching together certain Schroeder-Bernstein style maps. It's also shown in the paper that for every $n > 1$ and cardinal $\kappa$, there exists a linear order $X$ of size $\kappa$ such that $X^n \cong X$. There are actually many such orders, with diverse structural properties.

The result surprised me. In the majority of cases when one is able to find an infinite structure $X$ (e.g. group, topological space, graph, Boolean algebra) that is isomorphic to its cube, it is possible to find such a structure such that $X \not\cong X^2$. In the rare cases when it is possible to prove $X^3 \cong X \implies X^2 \cong X$ it is usually possible to prove the significantly stronger implication $A\times B\times X \cong X \implies B \times X \cong X$, which is false for linear orders. In fact it is even possible to construct orders $A, X$ such that $A^2 \times X \cong X$ but $A \times X \not \cong X$. (The corresponding right-sided implication $X \times B \times A \cong X \implies X \times B \cong X$ is also false.)

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Let me say what I know about the history and status of Sierpinski's other questions that bof quoted in their answer.

"We do not know so far any example of two types $\varphi$ and $\psi$, such that $\varphi^2 = \psi^2$ but $\varphi^3 \neq \psi^3$, or types $\gamma$ and $\delta$ such that $\gamma^2 \neq \delta^2$ but $\gamma^3 = \delta^3$."

Sierpinski knew of examples of non-isomorphic orders $X, Y$ whose squares are isomorphic. These examples are due to A.C. Morel and were generalized by Sierpinski. They have the property that not only is $X^2 \cong Y^2$ but in fact $X^n \cong Y^n$ for all $n>1$. Sierpinski's first question is whether the former condition actually implies the latter, or if there exist two orders whose squares are isomorphic but cubes are not.

Morel's examples show that linear orders need not have unique $n$th roots, that is, that $X^n \cong Y^n \implies X \cong Y$ is false over the class of linear orders for any fixed $n > 1$. However, the examples do not prove the falsity of the implication $X^n \cong Y^n \implies X^k \cong Y^k$ for $1 < k < n$. Sierpinski's second question is whether there is a counterexample when $n = 3$ and $k = 2$. Both of these questions remain open, to my knowledge.

"We do not know whether there exist two different denumerable order types which are left-hand divisors of each other."

Sierpinski was aware of distinct uncountable orders that are left-hand divisors of each other. It turns out the uncountability is necessary. This falls out of some of the work in the paper. A proof is given in the final section. (Sierpinski used the traditional anti-lexicographical ordering on products, so left-handed divisor for him is right-handed divisor in the convention of the paper.)

"Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other."

As already mentioned, Sierpinski was aware of distinct orders $X_0, Y_0$ that are left-hand divisors of one another. He also knew of distinct orders $X_1, Y_1$ that are right-hand divisors of one another. It is natural to ask if there are distinct orders $X, Y$ that divide each other on both the left and right.

As bof points out in their answer, if there were an order $X$ isomorphic to $X^3$ but not $X^2$, then the pair $X, X^2$ would give a positive answer. There is no such $X$, however, the answer to Sierpinski's question is still positive. The construction requires some work, and I do not yet have a writeup ready. The fact that such orders exist in some sense says that the implication $X^3 \cong X \implies X^2 \cong X$, while true for linear orders, is close to being false.

Edit 2: After thinking through this again, I realized that my argument for the final case, when $\tau$ has both a left and right endpoint, contained an error. I've now provided a new, much simpler argument for that case using the Schroeder-Bernstein theorem for linear orders. Somewhat surprisingly, this new argument applies to orders of arbitrary size: it shows that if $\tau$ is any order (not necessarily countable) with both a left and right endpoint, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$.

The two-endpoint case seems special: the new argument does not seem to apply to uncountable $\tau$ with no endpoints, or with a single endpoint. So those cases remain open.

I've also tried to clean up my answer a bit, hopefully making it more readable, despite its length.

New argument: We can still show that $\tau^2 \cong \tau$ in this case, but by an entirely different argument. The lemma we will need is the Shroeder-Bernstein theorem for linear orders. It says: if $X$ and $Y$ are linear orders such that $X$ is isomorphic to an initial segment of $Y$, and $Y$ is isomorphic to a final segment of $X$, then in fact $X \cong Y$. The proof of the theorem is the same as the usual one; the hypotheses guarantee that the bijection constructed between $X$ and $Y$ is order-preserving.

So suppose that $\tau$ has both a left and right endpoint, and for some $n$, $\tau^n \cong \tau$. We show $\tau^2 \cong \tau$ by Schroeder-Bernstein. Since $\tau$ has a left endpoint, we see that, viewing $\tau^2$ as $\tau$-many copies of $\tau$, that $\tau^2$ contains an initial copy of $\tau$. That is, $\tau$ is isomorphic to an initial segment of $\tau^2$. We will be done if we can show that $\tau^2$ is isomorphic to a final segment of $\tau^n$, since $\tau^n \cong \tau$. Viewing $\tau^n$ as $\tau^{n-2}$ copies of $\tau^2$, and using the fact that $\tau^{n-2}$ has a right endpoint (as $\tau$ does), we see that $\tau^n$ has a final copy of $\tau^2$, that is, $\tau^2$ maps isomorphically onto a final segment of $\tau^n$, as desired.

In no part of this argument did we need that $\tau$ was countable, so that indeed if $\tau$ is an order of arbitrary size with both endpoints, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$. Unfortunately, it seems impossible to use a similar argument in the other cases, since to apply Schroeder-Bernstein, we really need that $\tau$ has both endpoints.

Edit 2: After thinking through this again, I realized that my argument for the final case, when $\tau$ has both a left and right endpoint, contained an error. I've given a new, simpler argument for that case, using Lindenbaum's version of the Schroeder-Bernstein theorem for linear orders. The argument applies to orders of arbitrary size: it shows that if $\tau$ is any order (not necessarily countable) with both a left and right endpoint, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$.

The two-endpoint case seems special: the argument does not seem to apply to uncountable $\tau$ with no endpoints, or with a single endpoint. So those cases remain open.

New argument: We can still show that $\tau^2 \cong \tau$ in this case, but by an entirely different argument. The lemma we will need is Lindenbaum's version of the Schroeder-Bernstein theorem for linear orders. It says: if $X$ and $Y$ are linear orders such that $X$ is isomorphic to an initial segment of $Y$, and $Y$ is isomorphic to a final segment of $X$, then in fact $X \cong Y$. The proof of the theorem is the same as the usual one; the hypotheses guarantee that the bijection constructed between $X$ and $Y$ is order-preserving.

So suppose that $\tau$ has both a left and right endpoint, and for some $n$, $\tau^n \cong \tau$. We show $\tau^2 \cong \tau$ by Lindenbaum. Since $\tau$ has a left endpoint, we see that, viewing $\tau^2$ as $\tau$-many copies of $\tau$, that $\tau^2$ contains an initial copy of $\tau$. That is, $\tau$ is isomorphic to an initial segment of $\tau^2$. We will be done if we can show that $\tau^2$ is isomorphic to a final segment of $\tau^n$, since $\tau^n \cong \tau$. Viewing $\tau^n$ as $\tau^{n-2}$ copies of $\tau^2$, and using the fact that $\tau^{n-2}$ has a right endpoint (as $\tau$ does), we see that $\tau^n$ has a final copy of $\tau^2$, that is, $\tau^2$ maps isomorphically onto a final segment of $\tau^n$, as desired.

In no part of this argument did we need that $\tau$ was countable, so that indeed if $\tau$ is an order of arbitrary size with both endpoints, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$. Unfortunately, it seems impossible to use a similar argument in the other cases, since to apply Lindenbaum, we really need that $\tau$ has both endpoints.

Edit: As Joel pointed out in the comments, in the previous version of my answer, I reversed the usual meaning of $\times$ for linear orders. I've now (hopefully) corrected all instances of the reversal.

For countable $\tau$, I believe the answer is no. Joel Hamkins has commented on your Math.SE version of this question that from this we can get a negative answer for general $\tau$ by a forcing argument, giving a complete answer to your question. Joel, if you read this, it'd be great if you could give an explanation of your comment.

So assume that $\tau$ is a countable linear order, and that for some $n$ we have $\tau^n \cong \tau$. We show that in fact $\tau^2 \cong \tau$ (and so $\tau^m \cong \tau$ for every $m$). There are four possibilities: either $\tau$ has a top element (but no bottom), a bottom element (but no top), both a top and bottom element, or neither. For now let's assume we are in the last case, which is the simplest. The other cases are quite similar, and I'll describe how to handle them at the end.

The idea is to decompose $\tau$ in such a way that it is easy to see that $\tau^2 \cong \tau$. Forgetting $\tau$ for a moment, let's describe the kind of decomposition we are aiming for. First, decompose the set of rationals $\eta$ into countably many disjoint copies of itself. That is, write $\eta = \bigcup_{i \in \mathbb{N}} \eta_i$, where each $\eta_i \subseteq \eta$ is dense and for $i \neq j$ we have $\eta_i \cap \eta_j = \emptyset$. I'll refer to the elements of $\eta_i$ as $i$-points. One may think of $i$-points as being colored by the color $i$.

Now, suppose for each $i \in \mathbb{N}$ we have some countable linear order $L_i$. We may form a new order $L$ by replacing (in $\eta$), for each $i \in \mathbb{N}$, each $i$-point with a copy of $L_i$. Somewhat informally, I'll write \begin{equation} L = \bigcup_{i \in \mathbb{N}} L_i \times \eta_i. \end{equation} The claim is that $L^2 \cong L$. Indeed, if $\beta$ is any countable order, we have $L \times \beta \cong L$. For one may think of $L \times \beta$ as being formed in two steps: first, replace each point in $\beta$ with copy of $\eta$ to form $\eta \times \beta$, remembering our coloring of $\eta$. Then of course $\eta \times \beta$ is isomorphic to $\eta$, and moreover, for each $i$ the collection of $i$-points in $\eta \times \beta$ is dense. Thus there is an isomorphism between $\eta \times \beta$ and $\eta$ that sends $i$-points to $i$-points. (I am using here the fact that if $X$ and $Y$ are countable dense orders, each decomposed into countably many disjoint dense sets, $X = \bigcup_i X_i$, $Y = \bigcup_i Y_i$, then there is an isomorphism of $X$ and $Y$ that sends $X_i$ onto $Y_i$ for every $i$.) Since $L \times \beta$ is obtained by replacing each $i$-point in $\eta \times \beta$ with a copy of $L_i$, we see that $L \times \beta \cong L$, as claimed. Notice that the same argument works if some (or even all but finitely many) of the $L_i$ are empty.

Our goal then is to decompose $\tau$ similarly as $\bigcup_i L_i \times \eta_i$ for some collection of countable order types $L_i$, for then by above we have $\tau^2 \cong \tau$. By assumption, we have $\tau \cong \tau^n$. Let us instead write $\tau \cong \tau \times \beta$, where $\beta = \tau^{n-1}$. For us, the form of $\beta$ is irrelevant except that it is countable and, like $\tau$, has neither a top nor bottom element.

Now, the fact that $\tau \cong \tau \times \beta$, means that $\tau$ may be split up into $\beta$-many intervals, each of order type $\tau$ (where "$\beta$-many" is informal but hopefully clear). Each of these $\beta$-many copies of $\tau$ may in turn be split into $\beta$-many copies of $\tau$, and so on. This is as much to say that to every finite sequence $s = \langle x_1, x_2, \ldots, x_n \rangle$ of elements of $\beta$, we may associate an interval $I_s \cong \tau$. Namely, $I_s$ is the $x_n$-th copy of $\tau$ within the $x_{n-1}$-th copy of $\tau$ $\ldots$ within the $x_1$-st copy of $\tau$. We also have an $f_s: \tau \rightarrow I_s$ witnessing the isomorphism.

Notice that in iteratively splitting $\tau$ into smaller and smaller copies of itself, we may choose the $I_s$ and $f_s$ in such a way that the maps $f_s$ commute, that is, so that if $s^{\frown}t$ is the sequence obtained by concatenating the sequences $s$ and $t$, then $f_{s^{\frown}t} = f_{s} \circ f_{t}$ (where, on the right, "$f_s$" really means "$f_s \upharpoonright I_t$"). This I think is really the natural thing to do, but to be clear let me say what I mean. At the first stage, for every $x \in \beta$ we have an interval $I_{\langle x \rangle}$ of order type $\tau$ and an isomorphism $f_{\langle x \rangle}: \tau \rightarrow I_x$. For every $y \in \beta$, let $I_{\langle x, y \rangle}$ be the image of $I_{\langle y \rangle}$ under $f_{\langle x \rangle}$. Then naturally we obtain the isomorphism $f_{\langle x, y \rangle} = f_{\langle x \rangle} \circ f_{\langle y \rangle}$ from $\tau$ onto $I_{\langle x, y \rangle}$. And so on for sequences of length longer than 2. Note that if $s$ extends $t$, then $I_s$ is a proper subinterval of $I_t$.

Now, having associated to every finite sequence $s \in \beta^{<\omega}$ an interval $I_s$, we may associate to every infinite sequence $r \in \beta^{\omega}$ the interval $I_r = \bigcap_{n} I_{r \upharpoonright n}$ obtained by taking the natural nested intersection. The collection of $I_r$ for $r \in \beta^{\omega}$ is a covering of $\tau$ by nonintersecting intervals. Our construction guarantees that $I_r$ lies to the left of $I_s$ in $\tau$ if and only if $r < s$ in the lexicographical ordering of $\beta^{\omega}$. Note that for a fixed $r$, it may be that $I_r$ contains many points, a single point, or no points at all. Indeed, since $\tau$ itself is countable, $I_r$ will be empty for all but countably many $r$. The relevant observation for our purposes is that, if we view $I_r$ itself as a linear order, then for many densely $s \in \beta^{\omega}$ we have $I_r \cong I_s$. To see this, let us introduce an equivalence relation on the space $\beta^{\omega}$: say that $r \sim s$ iff $r$ and $s$ share a tail-sequence (not necessarily beginning at the same coordinate). That is: \begin{equation} r \sim s \leftrightarrow \textrm{there exists $m, n$ such that for every $i$ we have $r(m+i) = s(n+i)$}, \end{equation} where $r(k)$ means the $k$th entry in the sequence $r$, etc. This is clearly an equivalence relation. The claim is that if $r \sim s$, then $I_r \cong I_s$. For indeed, if $r \sim s$, then there are finite sequence $r', s' \in \beta^{<\omega}$ and an infinite sequence $x \in \beta^{\omega}$ such that $r = r'^{\frown}x$ and $s=s'^{\frown}x$. The intuitive reason that $I_r \cong I_s$ is that, in order to construct these intervals, we first move to $I_{r'}$ and $I_{s'}$, which are both just copies of $\tau$. Then, in each interval, we travel down the precisely corresponding nested sequence of intervals (represented by $x$) to obtain $I_r$ and $I_s$, which must therefore be isomorphic. More formally, we see that by the way we defined $f_{r'}$ and $f_{s'}$, we have $f_{s'} \circ f_{r'}^{-1}$ maps $I_r$ isomorphically onto $I_s$, establishing the claim.

For $s \in \beta^{\omega}$, let $\bar{s}$ denote its equivalence class. It is easy to see that if $s_1, s_2 \in \bar{s}$ and $s_1 < s_2$ (where again $<$ is the lexicographical ordering on $\beta^{\omega}$), then there are $r, r', r'' \in \bar{s}$ such that $r < s_1 < r' < s_2 < r''$ (we use here that $\beta$ has neither a top nor bottom element). It follows that $\bar{s}$ has the order type of the rationals. By what we just showed, for a fixed $\bar{s}$, every $I_r$ with $r \in \bar{s}$ has the same order type. Since only countably many of the $I_r$ are nonempty, we may enumerate the (distinct) classes $\overline{s_1}, \overline{s_2}, \ldots$, such that if $I_r$ is nonempty, there is $i$ such that $r \in \overline{s_i}$ (this list may be finite). Let $L_i$ be the shared order type of every $I_r$, $r \in \overline{s_i}$.

But now we have our desired decomposition. As already noted, every $\overline{s_i}$ has type $\eta$, and thus so does the subset $S \subseteq \beta^{\omega}$ given by $S = \bigcup_i \overline{s_i}$. We think of each $\overline{s}_i \subseteq S$ as our collection of $i$-points. But then by our choice of the $\overline{s_i}$, we have that $\tau$ may be formed by replacing each point $i$-point in $S$ by a copy of $L_i$. That is, $\tau = \bigcup_i \overline{s_i} \times L_i$. Then, by our argument given at the beginning, $\tau^2 \cong \tau$, as desired.

This takes care of the case when $\tau$ has neither a top nor bottom element. The other three cases are similar, but with complicating details. I will sketch the variations that arise. Hopefully I have not overlooked anything.

Suppose first that $\tau$ has a bottom element, but not a top one. In this case, the preliminary discussion given in paragraphs 3 and 4 is modified as follows. Instead of decomposing $\eta$ into countably many copies of itself, we decompose $1+\eta$ (that is, the order type of the rationals with a left endpoint added) as $1 + \eta_0 \, \, \cup \, \, \,\bigcup_{i>0} \eta_i$, where the $\eta_i$ are pairwise disjoint, and each is dense in $\eta$. This is the same as before, except that in our decomposition we have the distinguished set $1+\eta_0$ that includes the left endpoint, i.e. we color the left endpoint as a $0$-point. Suppose as before we are given a collection of countable orders $L_i$, where now we insist $L_0$ has a left endpoint. We may form $L$ by replacing every $i$-point with $L_i$, and write \begin{equation} (1) \, \, \, \, \, L = L_0 \times (1 + \eta_0) \, \, \cup \, \, \bigcup_{i>0} L_i \times \eta_i. \end{equation}

Then, we seek to use $\tau^n \cong \tau$ to decompose $\tau$ in the form of (1). Again we write $\tau$ as $\tau \times \beta$, where $\beta = \tau^{n-1}$, and note that $\beta$ has a left endpoint, which we label as 0. We again associate to every $s \in \beta^{<\omega}$ an interval $I_s$ and isomorphism $f_s$ of $I_s$ with $\tau$. By taking intersections we obtain the intervals $I_s$ for $s \in \beta^{\omega}$. Again if $s \sim r$, then $I_s \cong I_r$.

For each $s \in \beta^{\omega}$, the equivalence class $\bar{s}$ will again have the order type of the rationals, except for the equivalence class of the zero sequence $\mathbf{0} = \langle 0, 0, \ldots \rangle$. This class has order type $1 + \eta$, the left endpoint being $\mathbf{0}$ itself. Label this class as $\overline{s_0}$. Let $L_0$ be the shared order type of intervals associated to this class, and note that $L_0$ has a left endpoint. For $L_0$ is the order type of $I_{\mathbf{0}}$, which contains the left endpoint of $\tau$. We may enumerate the other classes corresponding to nonempty intervals as $\overline{s_1}, \overline{s_2}, \ldots$. Letting $L_i$ be the order type of the intervals associated to $\overline{s_i}$, we again see that $\tau$ may be written as $(L_0 \times \overline{s_0}) \cup \bigcup_{i>0} L_i \times \overline{s_i}$. Since $\overline{s_0}$ has type $1+\eta$ and all other $\overline{s_i}$ have type $\eta$, this is a decomposition of the desired form (1).

The case when $\tau$ has a top element but no bottom element is symmetric. Finally, suppose $\tau$ has both a top and bottom element. Briefly, the modifications in the argument are: begin by decomposing $1+\eta+1$ as $1 + \eta_0 \, \cup \eta_1 + 1 \, \cup \, \bigcup_{i > 1} \eta_i$ (so the left endpoint is a 0-point, the right endpoint is a 1-point). Suppose that $L_0$ is a countable order with a left endpoint, $L_1$ is a countable order with a right endpoint, for $i>1$ the $L_i$ are arbitrary countable orders. Then if $L$ is obtained by replacing each $i$-point in $1 + \eta + 1$ with a copy of $L_i$, we have that $L \times \beta \cong L$ for any countable $\beta$ also with a top and bottom point. In particular, $L^2 = L$. Toward decomposing $\tau$ in the form of $L$, we write $\tau = \tau \times \beta$ where $\beta = \tau^{n-1}$ has a bottom point 0 and a top point 1. Our classes $\bar{s}$ for $s \in \beta^{\omega}$ will all have order type $\eta$, except for the class of $\overline{\mathbf{0}}$, which has type $1+\eta$, and the class $\overline{\mathbf{1}}$, which has type $\eta+1$. Letting $L_0$ be the order type associated to $\overline{\mathbf{0}}$, we see that $L_0$ has a left endpoint. Similarly, $L_1$ has a right endpoint, where $L_1$ is the type associated the $\overline{\mathbf{1}}$. Labeling the other classes corresponding to nonempty order intervals as $\overline{s_2}, \overline{s_3}, \ldots$ we similarly obtain the desired decomposition of $\tau$ in the form of $L$.

Edit 2: After thinking through this again, I realized that my argument for the final case, when $\tau$ has both a left and right endpoint, contained an error. I've now provided a new, much simpler argument for that case using the Schroeder-Bernstein theorem for linear orders. Somewhat surprisingly, this new argument applies to orders of arbitrary size: it shows that if $\tau$ is any order (not necessarily countable) with both a left and right endpoint, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$.

The two-endpoint case seems special: the new argument does not seem to apply to uncountable $\tau$ with no endpoints, or with a single endpoint. So those cases remain open.

I've also tried to clean up my answer a bit, hopefully making it more readable, despite its length.

--

Edit: As Joel pointed out in the comments, in the previous version of my answer, I reversed the usual meaning of $\times$ for linear orders. I've now hopefully corrected all instances of the reversal.

For countable $\tau$, I believe the answer is no. Joel Hamkins has commented on your Math.SE version of this question that from this we can get a negative answer for general $\tau$ by a forcing argument, giving a complete answer to your question. Joel, if you read this, it'd be great if you could give an explanation of your comment. (Update: see Joel's comments and the edit above.)

Theorem. If $\tau$ is a countable linear order, and for some $n$ we have $\tau^n \cong \tau$, then $\tau^2 \cong \tau$.

Proof. There are four possibilities: either $\tau$ has no endpoints, $\tau$ has a left endpoint but not a right one, $\tau$ has a right endpoint but not a left one, or $\tau$ has both a left and right endpoint.

Case 1: Assume first that $\tau$ has no endpoints. We'll handle the other cases in turn.

The idea is to use the hypothesis $\tau^n \cong \tau$ to decompose $\tau$ in a certain way, and then use this decomposition to show $\tau^2 \cong \tau$. Forgetting $\tau$ for a moment, let's describe the form of the decomposition we are aiming for. Let $\eta$ denote the set of rationals. Partition $\eta$ into countably many disjoint dense sets, that is, find subsets $\eta_i \subseteq \eta$ such that each $\eta_i$ is dense, $\eta = \bigcup_{i \in \mathbb{N}} \eta_i$, and for $i \neq j$ we have $\eta_i \cap \eta_j = \emptyset$. I'll refer to the elements of $\eta_i$ as $i$-points. We think of $i$-points as being colored by the color $i$.

Now, suppose for each $i \in \mathbb{N}$ we are given some countable linear order $L_i$. We may form a new order $L$ by replacing, for each $i \in \mathbb{N}$, each $i$-point in $\eta$ with a copy of $L_i$. Somewhat informally, I'll write \begin{equation} L = \bigcup_{i \in \mathbb{N}} L_i \times \eta_i. \end{equation} Notice that the three examples you gave of infinite orders satisfying $\tau^2 \cong \tau$ are all of the form of $L$. To get $\eta$ for example, let $L_i = 1$ for every $i$. The orders $\omega \times \eta$ and $\omega^2 \times \eta$ are respectively obtained by letting $L_i = \omega$ for every $i$, and $L_i = \omega^2$ for every $i$. A more complicated order of this form is obtained by letting $L_0 = \omega$, and $L_i = \omega^2$ for $i > 0$. One may think of this order as $\omega \times \eta$ interspersed with $\omega^2 \times \eta$.

Lemma. For $L$ of this form, we always have $L^2 \cong L$. Indeed, if $\beta$ is any countable order, we have $L \times \beta \cong L$.

Proof. One may think of $L \times \beta$ as being formed in two steps. First, replace each point in $\beta$ with copy of $\eta$ to form $\eta \times \beta$. Since each copy of $\eta$ is partitioned into $i$-points, we may think of $\eta \times \beta$ as also being partitioned into $i$-points. Then for each $i$, the collection of $i$-points is dense in $\eta \times \beta$. Now, since $\beta$ is countable, $\eta \times \beta$ is isomorphic to $\eta$. Moreover, there is an isomorphism between $\eta \times \beta$ and $\eta$ that sends $i$-points to $i$-points. (I am using here a stronger form of Cantor's theorem that all countable dense orders without endpoints are isomorphic to $\eta$. Namely, if $X$ and $Y$ both have order type $\eta$, and each is decomposed into countably many disjoint dense sets, $X = \bigcup_i X_i$, $Y = \bigcup_i Y_i$, then there is an isomorphism of $X$ and $Y$ that sends $X_i$ onto $Y_i$ for every $i$.) Now, $L \times \beta$ is formed by replacing every $i$-point in $\eta \times \beta$ with a copy of $L_i$. Since $L$ is formed by replacing every $i$-point in $\eta$ with a copy of $L_i$, and there is an isomorphism between $\eta$ and $\eta \times \beta$ that respects our coloring, we must have $L \times \beta \cong L$. This proves the lemma. Notice that the same argument works if some (or even all but finitely many) of the $L_i$ are empty.

Our goal, then, is to show that $\tau^n \cong \tau$ implies that $\tau$ may be decomposed into the form of $L$; that is, $\tau = \bigcup_i L_i \times \eta_i$ for some collection of countable order types $L_i$. Then by the claim we will have $\tau^2 \cong \tau$. Let us write our hypothesis as $\tau \cong \tau \times \beta$, where $\beta = \tau^{n-1}$. For us, the form of $\beta$ is irrelevant except that it is countable and, like $\tau$, has neither a left nor right endpoint.

The fact that $\tau \cong \tau \times \beta$, means that $\tau$ may be split up into $\beta$-many intervals, each of order type $\tau$. Each of these $\beta$-many copies of $\tau$ may in turn be split into $\beta$-many copies of $\tau$, and so on. This is as much to say that to every finite sequence $s = \langle x_1, x_2, \ldots, x_n \rangle$ of elements of $\beta$, we may associate an interval $I_s$ that is isomorphic to $\tau$. Namely, $I_s$ is the $x_n$-th copy of $\tau$ within the $x_{n-1}$-th copy of $\tau$ $\ldots$ within the $x_1$-st copy of $\tau$. Since every $I_s$ is of order type $\tau$, we also have some $f_s: \tau \rightarrow I_s$ witnessing the isomorphism.

In iteratively splitting $\tau$ into smaller and smaller copies of itself, we may choose the $I_s$ and $f_s$ in such a way that the maps $f_s$ respect concatenation, that is, so that if $s^{\frown}t$ is the sequence obtained by concatenating the sequences $s$ and $t$, then $f_{s^{\frown}t} = f_{s} \circ f_{t}$ (where, on the right, $f_s$ really means $f_s \upharpoonright I_t$). Here's how to do this. At the first stage, for every $x \in \beta$ we have an interval $I_{\langle x \rangle}$ of order type $\tau$ and an isomorphism $f_{\langle x \rangle}: \tau \rightarrow I_{\langle x \rangle}$. We define $I_s$ and $f_s$ for all longer $s$ in terms of these first-level $f_{\langle x \rangle}$ and $I_{\langle x \rangle}$. At the second stage, for every $y \in \beta$, let $I_{\langle x, y \rangle}$ be the image of $I_{\langle y \rangle}$ under $f_{\langle x \rangle}$. Then we simply define $f_{\langle x, y \rangle}$ to be$ f_{\langle x \rangle} \circ f_{\langle y \rangle}$, which by our choice of $I_{\langle x, y \rangle}$ is an isomorphism $\tau$ onto $I_{\langle x, y \rangle}$. For every $z \in \beta$, let $I_{\langle x, y, z \rangle} = f_{\langle z \rangle}[I_{\langle x, y \rangle}]$, and and $f_{\langle x, y, z \rangle} = f_{\langle x \rangle} \circ f_{\langle y \rangle} \circ f_{\langle z \rangle}$. And so on for longer sequences. The $f_s$ so defined clearly respect concatenation. Note that if $s$ extends $t$, then $I_s$ is a proper subinterval of $I_t$.

It may be helpful to illustrate this construction with an example. If, say, $X = [0, 1)$, then $X \cong X \times 2$, as witnessed by splitting up $X$ as $[0, \frac{1}{2}) \cup [\frac{1}{2}, 1)$. Here, $2=\{0, 1\}$ is playing the role of $\beta$, and we have $I_0 = [0, \frac{1}{2})$, and $I_1=[\frac{1}{2}, 1)$. Let $f_0(x) = \frac{1}{2}x$ be our isomorphism from $X$ onto $I_0$ and $f_1(x) = \frac{1}{2}x + \frac{1}{2}$ be our isomorphism from $X$ onto $I_1$. Then, for example, $I_{01} = f_0[I_1] = [\frac{1}{4}, \frac{1}{2})$ and $f_{01} = f_0 \circ f_1 = \frac{1}{2}(\frac{1}{2} x + \frac{1}{2}) = \frac{1}{4}x + \frac{1}{4}$ is our isomorphism from $X$ onto $I_{01}$. By taking longer compositions of $f_0$ and $f_1$, we obtain $I_s$ and $f_s$ for every $s \in 2^{<\omega}$.

Now, back in our setting, having associated to every finite sequence $s \in \beta^{<\omega}$ an interval $I_s$, we may associate to every infinite sequence $r \in \beta^{\omega}$ the interval $I_r = \bigcap_{n} I_{r \upharpoonright n}$ obtained by taking the natural nested intersection. The collection of $I_r$ for $r \in \beta^{\omega}$ is a covering of $\tau$ by nonintersecting intervals. Let us view $\beta^{\omega}$ also as a linear order, under the lexicographical ordering of sequences. Then our construction guarantees that $I_r$ lies to the left of $I_{r'}$ in $\tau$ if and only if $r <_{lex} r'$ in $\beta^{\omega}$. Indeed, $\tau$ is recovered from $\beta^{\omega}$ by replacing every $r \in \beta^{\omega}$ by the corresponding interval $I_r$. Note, however, that for a fixed $r$, it may be that $I_r$ contains many points, a single point, or no points at all. In fact, since $\tau$ itself is countable, $I_r$ will be empty for all but countably many $r$. 

The relevant observation is that, if we view the interval $I_r$ as a linear order, then for densely many $s \in \beta^{\omega}$ we have $I_r \cong I_s$. To see this, let us introduce an equivalence relation on the space $\beta^{\omega}$: say that $r \sim s$ iff $r$ and $s$ share a tail-sequence (not necessarily beginning at the same coordinate). That is: \begin{equation} r \sim s \leftrightarrow \textrm{there exists $m, n$ such that for every $i$ we have $r(m+i) = s(n+i)$}, \end{equation} where $r(k)$ means the $k$th entry in the sequence $r$, etc. This is clearly an equivalence relation. Notice that since $\beta$ is countable, every $\sim$-equivalence class is countable.

Claim. If $r \sim s$, then $I_r \cong I_s$.

Proof. If $r \sim s$, then there are finite sequence $r', s' \in \beta^{<\omega}$ and an infinite sequence $x \in \beta^{\omega}$ such that $r = r'^{\frown}x$ and $s=s'^{\frown}x$. The intuitive reason that $I_r \cong I_s$ is that, in order to construct these intervals, we first move to $I_{r'}$ and $I_{s'}$, which are both just copies of $\tau$. Then, in each interval, we travel down the precisely corresponding nested sequence of intervals (represented by $x$) to obtain $I_r$ and $I_s$, which must therefore be isomorphic. More formally, we see that by the way we defined $f_{r'}$ and $f_{s'}$, we have $f_{s'} \circ f_{r'}^{-1}$ maps $I_r$ isomorphically onto $I_s$, establishing the claim.

For $s \in \beta^{\omega}$, let $\bar{s}$ denote its equivalence class. It is easy to see that if $s_1, s_2 \in \bar{s}$ and $s_1 < s_2$ in the lexicographical order, then there are $r, r', r'' \in \bar{s}$ such that $r < s_1 < r' < s_2 < r''$ (we use here that $\beta$ has neither a top nor bottom element). Since $\bar{s}$ is countable, it follows that $\bar{s}$ (viewed as a suborder of $\beta^{\omega}$) has order type $\eta$. In fact, it's similarly easy to see that $\bar{s}$ is not only dense in its order type, but is actually dense in $\beta^{\omega}$.

Now by the claim, for a fixed $\bar{s}$, every $I_r$ with $r \in \bar{s}$ is isomorphic. Since only countably many of the $I_r$ are nonempty, we may enumerate the (distinct) classes $\overline{s_1}, \overline{s_2}, \ldots$, such that if $I_r$ is nonempty, there is $i$ such that $r \in \overline{s_i}$ (this list may even be finite). Let $L_i$ be the shared order type of every $I_r$, $r \in \overline{s_i}$.

But now we have our desired decomposition. As already noted, every $\overline{s_i}$ has type $\eta$, and is dense in $\beta^{\omega}$. Thus the subset $S \subseteq \beta^{\omega}$ given by $S = \bigcup_i \overline{s_i}$ also has order type $\eta$. We think of each $\overline{s}_i \subseteq S$ as our collection of $i$-points. But then since the $\overline{s_i}$ range over all nonempty intervals $I_r$, we have that $\tau$ may be formed by replacing each $i$-point in $S$ by a copy of $L_i$. That is, $\tau = \bigcup_i \overline{s_i} \times L_i$. This is a decomposition of the sought after form. By our lemma at the beginning, we have $\tau^2 \cong \tau$, as desired.

This takes care of the case when $\tau$ has neither a left nor right endpoint. The case when $\tau$ has a single endpoint is similar, but with complicating details.

Case 2 (sketch): Suppose now that $\tau$ has a left endpoint, but not a right one. The case where $\tau$ has a right endpoint but not a left one is symmetric.

We seek to adapt the proof of case 1. The preliminary discussion preceding the lemma is modified as follows. Instead of decomposing $\eta$ into countably many copies of itself, we decompose $1+\eta$ (that is, the order type of the rationals with a left endpoint added) as $1 + \eta_0 \, \, \cup \, \, \,\bigcup_{i>0} \eta_i$, where the $\eta_i$ are pairwise disjoint, and each is dense in $\eta$. This is the same as before, except that in our decomposition we have the distinguished set $1+\eta_0$ that includes the left endpoint, i.e. we color the left endpoint as a $0$-point. Suppose as before we are given a collection of countable orders $L_i$, where now we insist $L_0$ has a left endpoint. We may form $L$ by replacing every $i$-point with $L_i$, and write \begin{equation} L = L_0 \times (1 + \eta_0) \, \, \cup \, \, \bigcup_{i>0} L_i \times \eta_i. \tag{$\star$} \end{equation}

Then, we seek to use $\tau^n \cong \tau$ to decompose $\tau$ in the form of ($\star$). Again we write $\tau$ as $\tau \times \beta$, where $\beta = \tau^{n-1}$, and note that $\beta$ has a left endpoint, which we label as 0. We again associate to every $s \in \beta^{<\omega}$ an interval $I_s$ and isomorphism $f_s$ of $I_s$ with $\tau$. By taking intersections we obtain the intervals $I_s$ for $s \in \beta^{\omega}$. Again if $s \sim r$, then $I_s \cong I_r$.

For each $s \in \beta^{\omega}$, the equivalence class $\bar{s}$ will again have the order type of the rationals, except for the equivalence class of the zero sequence $\mathbf{0} = \langle 0, 0, \ldots \rangle$. This class has order type $1 + \eta$, the left endpoint being $\mathbf{0}$ itself. Label this class as $\overline{s_0}$. Let $L_0$ be the shared order type of intervals associated to this class, and note that $L_0$ has a left endpoint. For $L_0$ is the order type of $I_{\mathbf{0}}$, which contains the left endpoint of $\tau$. We may enumerate the other classes corresponding to nonempty intervals as $\overline{s_1}, \overline{s_2}, \ldots$. Letting $L_i$ be the order type of the intervals associated to $\overline{s_i}$, we again see that $\tau$ may be written as $(L_0 \times \overline{s_0}) \cup \bigcup_{i>0} L_i \times \overline{s_i}$. Since $\overline{s_0}$ has type $1+\eta$ and all other $\overline{s_i}$ have type $\eta$, this is a decomposition of the desired form ($\star$).

This completes the proof of case 2. As noted, the argument for when $\tau$ has a right endpoint but not a left one is symmetric.

Case 3: Now suppose that $\tau$ has both a left and right endpoint.

Update: I originally gave an analogous argument for this case, but there is a detail that prevents the argument from succeeding. Here is the error. In the first case, when $\tau$ had no endpoints, we used the fact that for an arbitrary countable $\beta$, we have $\eta \times \beta \cong \eta$. In the second, we noted that as long as $\beta$ has a left endpoint, $(1+\eta) \times \beta \cong 1 + \eta$. For this final case, the analogous claim would be that if $\beta$ has both a left and right endpoint, then $(1+\eta+1)\times \beta \cong 1+\eta+1$. While this is true if $\beta$ is itself $1+\eta+1$, it is false in general; indeed it is false for any $\beta$ that is not dense. In particular, while the other details of the argument go through, it does not follow that $\tau^2 \cong \tau$, and I do not see how this argument can be fixed.

New argument: We can still show that $\tau^2 \cong \tau$ in this case, but by an entirely different argument. The lemma we will need is the Shroeder-Bernstein theorem for linear orders. It says: if $X$ and $Y$ are linear orders such that $X$ is isomorphic to an initial segment of $Y$, and $Y$ is isomorphic to a final segment of $X$, then in fact $X \cong Y$. The proof of the theorem is the same as the usual one; the hypotheses guarantee that the bijection constructed between $X$ and $Y$ is order-preserving.

So suppose that $\tau$ has both a left and right endpoint, and for some $n$, $\tau^n \cong \tau$. We show $\tau^2 \cong \tau$ by Schroeder-Bernstein. Since $\tau$ has a left endpoint, we see that, viewing $\tau^2$ as $\tau$-many copies of $\tau$, that $\tau^2$ contains an initial copy of $\tau$. That is, $\tau$ is isomorphic to an initial segment of $\tau^2$. We will be done if we can show that $\tau^2$ is isomorphic to a final segment of $\tau^n$, since $\tau^n \cong \tau$. Viewing $\tau^n$ as $\tau^{n-2}$ copies of $\tau^2$, and using the fact that $\tau^{n-2}$ has a right endpoint (as $\tau$ does), we see that $\tau^n$ has a final copy of $\tau^2$, that is, $\tau^2$ maps isomorphically onto a final segment of $\tau^n$, as desired.

In no part of this argument did we need that $\tau$ was countable, so that indeed if $\tau$ is an order of arbitrary size with both endpoints, then $\tau^n \cong \tau$ implies $\tau^2 \cong \tau$. Unfortunately, it seems impossible to use a similar argument in the other cases, since to apply Schroeder-Bernstein, we really need that $\tau$ has both endpoints.