Suppose that I have structures $M \preceq M'$ (in some first-order language). I have a set $A$, with $M \subseteq A \subseteq M'$, and an automorphism $f$ of $M'$. Is it is always possible to find an $M''$, with $M \preceq M'' \preceq M'$, and $A \subseteq M''$, such that $f$ restricts to an automorphism of $M''$? If so, and supposing that $M$ and $A$ are countable, can I also arrange for $M''$ to be countable?
Thanks!
Consider $M_0$ to be the elementary submodel generated by $A\cup f[A]\cup f^{-1}[A]$. Now define by induction $M_{n+1}$ to be the elementary submodel generated by $M_n\cup f[M_n]\cup f^{-1}[M_n]$.
Let $M''$ be the union of the $M_n$'s. Increasing unions of elementary submodels is an elementary submodel, and it is not hard to see that if $m\in M''$ then $f(m)\in M''$ as well, and if $f(m)\in M''$ then $m\in M''$.
Furthermore by the usual Lowenheim-Skolem arguments we have that $|M_n|=|A|+\aleph_0$ and therefore if $A$ is countable, so is $M''$.