[This partial answer was posted Nov. 19 to Math Stack Exchange but seems to have been ignored there. It seems to be a more concise version of the easy case (no endpoints) of Garrett Ervin's answer.]
Such questions have been considered in the past. From W. Sierpiński's Cardinal and Ordinal Numbers, second edition revised, Warszawa 1965, p. 235: "We do not know so far any example of two types $\varphi$ and $\psi$, such that $\varphi^2=\psi^2$ but $\varphi^3\ne\psi^3$, or types $\gamma$ and $\delta$ such that $\gamma^2\ne\delta^2$ but $\gamma^3=\delta^3$. Neither do we know any type $\alpha$ such that $\alpha^2\ne\alpha^3=\alpha$." Also, from p. 254: "We do not know whether there exist two different denumerable order types which are left-hand divisors of each other. Neither do we know whether there exist two different order types which are both left-hand and right-hand divisors of each other." Of course, if $\tau^n=\tau$ for some integer $n\gt2$, then $\tau^2$ and $\tau=\tau^2\tau^{n-2}=\tau^{n-2}\tau^2$ are both left-hand and right-hand divisors of each other.
For what it's worth, here is a partial answer to your question, for a very special class of order types. By "order type" I mean linear order type. An order type $\xi$ is said to have a "first element" if it's the type of an ordered set with a first element, i.e., if $\xi=1+\psi$ for some $\psi$; the same goes for "last element".
Proposition. If $\alpha$ is a countable order type, and if $\alpha\xi=\alpha$ for some order type $\xi$ with no first or last element, then $\alpha\beta=\alpha$ for every countable order type $\beta\ne0$.
Corollary. If $\tau$ is a countable order type with no first or last element, and if $\tau^n=\tau$ for some integer $n\gt1$, then $\tau^2=\tau$.
The corollary is obtained by setting $\alpha=\beta=\tau$ and $\xi=\tau^{n-1}$ in the proposition.
The proposition is proved by a modified form of Cantor's back-and-forth argument. Namely, let $A$ be an ordered set of type $\alpha=\alpha\xi$, and let $B$ be an ordered set of type $\alpha\beta=\alpha\xi\beta$. Since $A$ and $B$ are countable sets, let's fix an enumeration of each set.
An isomorphism between $A$ and $B$ will be constructed as the union of a chain of partial isomorphisms $f_k$ of the following form. The domain of $f_k$ is $I_1\cup I_2\cup\dots\cup I_k$, where $I_1,\dots,I_k$ are intervals in $A$ of order type $\alpha$; $I_1\lt\dots\lt I_k$; the interval in $A$ between $I_j$ and $I_{j+1}$ ($1\le j\lt k$), as well as the interval to the left of $I_1$ and the interval to the right of $I_k$, have order types which are nonzero right multiples of $\alpha$. The range of $f_k$ is $J_1\cup\dots\cup J_k$ where $J_1,\dots,J_k$ are intervals in $B$ of type $\alpha$, etc. etc. etc., and $f(I_1)=J_1,\dots,f(I_k)=J_k$.
Here is an informal description of the induction step. Suppose we have defined $f_k$ etc. as described above. Let $x$ be the first point in the enumeration of $A$ which is not in the domain of $f_k$, say $I_1\lt x\lt I_2$. The gap between $I_1$ and $I_2$ is of order type $\alpha\gamma$ for some order type $\gamma\gt0$. Since $\alpha\xi=\alpha$, we have $\alpha\gamma=\alpha\xi\gamma=\alpha\theta$ where $\theta=\xi\gamma$ is a nonzero order type with no first or last element. That is, the gap between $I_1$ and $I_2$ is the linear sum of a $\theta$-type set of intervals of type $\alpha$, and $x$ is in one of those intervals, let's call it $I$. Note that the gap between $I_1$ and $I$, as well as the gap between $I$ and $I_2$, has an order type which is a nonzero right multiple of $\alpha$. Choose an interval $J$ between $J_1$ and $J_2$ in the same way, except that there is no particular point to be covered. Extend $f_k$ to $f_{k+1}$ so that $I$ is mapped isomorphically onto $J$.
