An order type $\tau$ equal to its power $\tau^n, n>2$

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 $\alpha\beta$. 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 $f(I_1)=J_1,\dots,f(I_k)=J_k$.

This doesn't actually answer the question asked, admittedly, but I'm posting it as an answer since it does answer a related question.

We can ask more generally, can we find an order $\tau$ such that there exist distinct $n$ and $m$ such that $\tau^n=\tau^m$, other than those satisfying $\tau=\tau^2$? Note that if we have such a $\tau$, then either there exists some $n$ such that $\tau^n=\tau^{n+1}$, or by taking an appropriate power of $\tau$ we can find an order that actually answers the original question.

Well, as I said, I don't have an example of the latter, but I do have an example of the former. Let $\tau=\omega(\eta+1)$. Since $(\eta+1)\omega=\eta$, we get $\tau^2=\omega\eta$, and similarly $\tau^3=\omega\eta=\tau^2$. But $\omega(\eta+1)\ne\omega\eta$ as the former has a final segment isomorphic to $\omega$ and the latter doesn't (as one way to see why, note that any final segment contains a copy of $\omega2$).

So at least there exists $\tau$ with $\tau^3=\tau^2$ but not $\tau^2=\tau$, though that doesn't seem to yield an answer to the original question.