This question stems from a weakness in this recent answer of mine.
Question: Can $\mathsf{ZFC}$ prove, without invoking first-order model theory, that for every countably infinite structure $\mathcal{A}$ in a countable language there is an ultrafilter $\mathcal{U}$ on $\omega$ such that the ultrapower $\mathcal{A}^\mathcal{U}$ has a nontrivial automorphism?
"Obviously" the answer is yes! (... right?)
Note that this is trivial in $\mathsf{ZFC+CH}$: $\mathsf{ZFC}$ alone proves that any nontrivial ultrapower is countably saturated, which in the presence of $\mathsf{CH}$ lets us run a back-and-forth argument to show that $\mathcal{A}^\mathcal{U}$ is non-rigid whenever $\mathcal{A}$ is countable and $\mathcal{U}$ is a nonprincipal ultrafilter on $\omega$. Separately, note that if we drop both the model theory prohibition and the requirement that the ultrafilter be on $\omega$ a positive answer follows from the Keisler-Shelah theorem, but $(i)$ that doesn't count and $(ii)$ $\mathsf{ZFC}$ can't bring down KS to $\omega$ as one might expect, so it ultimately doesn't seem like a promising direction.
I strongly suspect that I'm missing a very simple argument, but at present I'm not seeing it.
Now asked at MO.