(All rings here are assumed to be commutative and unital)
I have a rather naive question (it is naive since in general, I expect the problem to be very hard).
Is it possible to determine when automorphisms of a ring preserve a module structure with respect to some other ring?
The same question in a bit more detail:
Let $R$ be some ring (for simplicity it may be $k$-algebra for some field $k$) and we have an action of another commutative unital ring ($k$-algebra) $A$ on the ring $R.$ Assume that $\phi$ is a ring automorphism of $R,$ when is it an automorphism of $A$-modules as well?
Maybe a reasonable answer could be achieved if we restrict to the case of finite-dimensional local $k$-algebra $A$ and finite-dimensional $k$-algebra $R$?
I am assuming that by action of $A$ on $R$ you mean that $R$ is an $A$-algebra. Such situation corresponds precisely to the datum of a ring (or $k$-algebra) homomorphism $\psi:A \rightarrow R$, which is determined by the rule $\psi(a)=a \cdot 1_R$. Conversely, the $A$-module action is determined by $\psi$ by the rule $a \cdot r:=\psi(a)r$, where the right-hand side operation is just multiplication in $R$.
Then the claim is that $R \stackrel{\phi}\rightarrow R$ is an $A$-module homomorphism if and only if $\psi(A)$ is fixed by $\phi$, i.e. $\phi(x)=x$ for every $x \in \psi(A)$:
1) On one hand, if this is true then $$\phi(a \cdot r)=\phi(\psi(a)r)=\phi(\psi(a))\phi(r)=\psi(a)\phi(r)=a \cdot \phi(r), \;\; a \in A, r \in R,$$ so $\phi$ is an $A$-module homomorphism.
2) Conversely, assuming that $\phi$ is an $A$-module homomorphism, we have for $a \in A$ $$\phi(\psi(a))=\phi(\psi(a)1)=\phi(a \cdot 1)=a \cdot \phi(1)=a \cdot 1=\psi(a),$$ showing that $\psi(A)$ is fixed by $\phi$.