I've been trying to come up with an example of two models $\mathcal{M,N}\models PA$ such that $\mathcal{M}\not\cong\mathcal{N}$, but $Aut(\mathcal{M})\cong Aut(\mathcal{N})$.
I know that for countable recursively saturated models this is still an open question (Automorphism Groups of Countable Arithmetically Saturated Models of PA, Schmerl, 2014), but I was wondering if there is an example for such models without the limitations of being countable and recursively saturated.
Any help would be appreciated!
There are a lot of rigid models of $\textsf{PA}$ -- models with no nontrivial automorphisms. For example, any finitely generated model of arithmetic is rigid (this is an easy application of Ehrenfeucht's Lemma, which says that if $a, b$ are in $\mathcal{M}$ are such that there is a Skolem term $t(x)$ with $\mathcal{M} \models t(a) = b$, then their types are different). So take $\mathbb{N}$ and any finitely generated elementary extension of $\mathbb{N}$ -- they will both be rigid but will not be isomorphic.