Elementary equivalence and isomorphism in some model of ZFC If two structures are isomorphic in some model of $\mathsf{ZFC}$ then they are elementary equivalent in this model and therefore in every model of $\mathsf{ZFC}$. If I am right, this is for instance used in the famous result by Ax and Kochen that two Henselian valued fields of same characteristic are elementary equivalent if and only if their residue fields and value groups that are elementary equivalent.
I am interested in the converse:

If two structures are elementary equivalent, does there exist a model of $\mathsf{ZFC}$ in which they are isomorphic?

In applications, this would be interesting in the following situation, for instance: Say we want to prove that two structures are isomorphic under $\mathsf{ZFC}$, but we can only show so far that they are elementary equivalent. Then we would at least know that it does not follow from $\mathsf{ZFC}$ that there is no isomorphism.
 A: As mentioned in the comments, the question is a bit imprecise. Let us make this a bit more precise.

Suppose $M,N\in V$ are elementarily equivalent structures. Is there a model $V'\supseteq V$ whose $V$ is a transitive submodel (i.e. the membership relation in $V$ is the restriction of the membership relation in $V'$ and no element of $V$ has more elements in $V'$ than it does in $V$), in which there is an isomorphism between $M$ and $N$?

Note that this is about as general as it makes sense to consider this question. If $V$ was not transitive in $V'$, then $M$ and $N$ might have more elements in $V'$, and the relations might be completely different, completely changing the structure (in particular, they could no longer be elementarily equivalent); in fact, if the signature $L$ has function symbols, then I think $M$ and $N$ could no longer be $L$-structures. And if $V$ was non-standard in $V'$ (i.e. the membership relation was not the restriction of the membership relation of $V'$), then this makes even less sense.
I believe the answer to this more precise question is no (in general), at least if you assume that $V$ and $V'$ are to be models of ZFC (I suspect the answer is the same for models of ZF or its reasonable fragment, but I'm not sure). More precisely, if $M\cong N$ in a larger model if and only if there is a back-and-forth system between $M$ and $N$, i.e. a family $\mathcal F$ of finite elementary partial functions $M\to N$ such that for every $f\in \mathcal F$, $m\in M,n\in N$, there is $f'\in \mathcal F$ such that $m$ is in the domain and $n$ is in the range of $f'$.
Now, given $M,N,V,V'$ as in the statement and $\mathcal F\in V$, we have that $\mathcal F$ is a back-and-forth system for $M$ and $N$ in $V$ if and only if it is that in $V'$: if you spell out all the definitions, everything is $\Delta_0$-definable, i.e. defined in terms of bounded quantifiers, and so absolute for transitive models (if you don't know this, I encourage you to try and work this out yourself, this is an easy exercise).
In particular, if $M$ and $N$ are back-and-forth equivalent in $V$, then we can find a $V'$ in which $M$ and $N$ are both countable, and then it's easy to check that there is an isomorphism (by a standard back-and-forth argument).
The converse is a bit more complicated. The reason is that the existence of a back-and-forth system for $M$ and $N$ clearly implies that $M$ and $N$ satisfy the same $L_{\infty,\omega}$-sentences in $V$ (this can be checked by straightforward induction), and the converse is also true: if two structures satisfy the same $L_{\infty,\omega}$-sentences, they is a back-and-forth system. In fact, we can write one sentence which expresses this (a so-called Scott sentence), i.e. given $M$, there is an $L_{\infty,\omega}$-sentence $\Phi$ such that $N\models\Phi$ iff it is back-and-forth equivalent to $M$.
Given this, if $M$ and $N$ are isomorphic in $V'$, they clearly satisfy the same $L_{\infty,\omega}$-sentences in $V'$. Now, every $L_{\infty,\omega}$-sentence in $V$ is still an $L_{\infty,\omega}$-sentence in $V'$ and its satisfaction is absolute (again, it's $\Delta_0$-definable). In particular, $N$ satisfies the Scott sentence for $M$ in $V$, so there is a back-and-forth system for $M$ and $N$ in $V$. (If $V$ was an inner model of $V'$, IIRC, it would follow that Scott sentences in $V$ and $V'$ are the same, but I'm not sure this is true if $V$ is only transitive in $V'$ which may have more ordinals.)
