**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.