TITLE: An order type $\tau$ equal to its power $\tau^n, n>2$
QUESTION [21 upvotes]: (This is a re-post of my old unanswered question from Math.SE)
For purposes of this question, let's concern ourselves only with linear (but not necessarily well-founded) order types.
Recall that:

$0, 1, 2, \dots$ — unique linear order types for each finite cardinality, that can be identified with natural numbers.
$\omega$ — the order type of $\mathbb{N}$ ordered by magnitude, the smallest infinite ordinal. 
$\eta$ — the dense countable order type of rational numbers $\mathbb{Q}$ ordered my their magnitude, which is also an order type of any dense denumerable linear order without first and last elements (e.g. the set of positive algebraic numbers).

The sum and product of order types are natural generalizations of (and are consistent with) the sum and product of ordinals, which we consider well-known. Let's agree that a positive integer power of an order type is just a syntactic shortcut for repeated multiplication.
There are some order types satisfying $\tau^2=\tau$, for example: $0,\,1,\,\eta,\,\omega\,\eta$ and $\omega^2\eta$.

Question: Is there a linear order type $\tau$ such that $\tau^2\ne\tau$, but $\tau^n=\tau$ for some integer $n>2$?

REPLY [6 votes]: [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 $J$.