Let $A,B,C$ be models and suppose that $A \cong B \preccurlyeq C$. What simple additional requirement is sufficient to entail that $A \preccurlyeq C$?
Notes:
- To see that this does not always follow, recall that $(\mathbb{N}\setminus\{0\}) \cong \mathbb{N} \preccurlyeq \mathbb{N}$ and yet $(\mathbb{N} \setminus\{0\}) \not\preccurlyeq \mathbb{N}$
- "$\cong$" denotes model isomorphism, "$\preccurlyeq$" denotes the elementary submodel relation.
Here's one possible answer: Model completeness of the theory $T$. Recall that $T$ is model complete if whenever $A$ and $B$ are models of $T$ and $A$ is a substructure of $B$, then $A$ is an elementary substructure of $B$.
In your case, if $T$ is model complete, $A\cong B$, $B\preccurlyeq C$, and $A$ is a substructure of $C$, then $A\preccurlyeq C$. Of course, this is a little silly: We didn't actually use $B$, already the fact that $A$ is a substructure of $C$ which is a model of $T$ was enough.
But if you ask for this conclusion to hold for all models of $T$, model completeness is a necessary hypothesis. To see this, suppose that $T$ is not model complete. Then I can find some a model $A$ which is a substructure of a model $C$, such that $A\not\preccurlyeq C$. But by compactness, there is an elementary extension $C\preccurlyeq C'$ and $B\cong A$ such that $B\preccurlyeq C'$. But then $A$ is a substructure of $C'$ and $A\not\preccurlyeq C'$, but $A\cong B\preccurlyeq C'$, contradicting our condition on $T$.
Edit: As pointed out below in the comments, the argument above only works when $T$ is complete (the step where an isomorphic copy of $A$ is elementarily embedded in an elementary extension of $C$ uses $A\equiv C$). What the proof above actually shows is that the following are equivalent: