TITLE: Is it possible to have two models $M$, $N$ and for each one an elementary embedding to the other one, but $M$, $N$ are not isomorphic?
QUESTION [9 upvotes]: Is it possible to have two models $M$ and $N$ of a theory $T$, an elementary embedding $f$ from $M$ to $N$, and also an elementary embedding $g$ from $N$ to $M$, but such that $M$ and $N$ are not isomorphic?
I myself think the answer is no, but I don't have any good reasons. I'm looking for an example.

REPLY [5 votes]: Here is an example with countable models, in fact, countable ordered sets. (By ordered set and order type I mean linearly ordered set and linear order type.)
Lemma 1. Let $\varphi,\psi$ be order types. If $(\omega^*+\omega)\varphi=(\omega^*+\omega)\psi,$ then $\varphi=\psi.$
Lemma 2. Let $M,N$ be ordered sets with order types $\operatorname{tp}M=(\omega^*+\omega)\varphi,\ \operatorname{tp}N=(\omega^*+\omega)\psi.$
If $\varphi\le\psi,$ then $M$ is elementarily embeddable in $N.$
Lemma 1 is more or less obvious. For Lemma 2, the method of Ehrenfeucht–Fraïssé games may be used to show that the natural embedding is elementary.
Now all we have to do is find countable order types $\varphi,\psi$ such that $\varphi\le\psi\le\varphi$ while $\varphi\ne\psi.$ For this we can take $\varphi=\eta$ and $\psi=1+\eta$ where $\eta$ is the order type of the rational numbers; or, if scattered orderings are preferred, $\varphi=\omega^*\omega$ and $\psi=1+\omega^*\omega$ will do.