Let $D$ be a division ring, and consider any matrix ring $R:=M_r(D)$. Of course, $R$ becomes a left $R$-Module via left multiplication. Let's call the latter $IR$ for distinction.
I wanted to prove that all maximal left submodules of $IR$ are conjugated by an automorphism. They take the form $\{M\mid V\subseteq \mathrm{Ker}\ M\}$ for a one-dimensional subspace $V\leq D^r$.
However, if we let $γ\in \mathrm{Aut}\ R\colon M\mapsto {}^SM := SMS^{-1}$, we note that $$ γ(NM) = {}^SNγ(M), $$ So $γ$ cannot be a homomorphism $IR\to IR$ of left $R$-Modules, since it does not commute with the left $R$-action.
This is confusing to me. Since left submodules correspond to left ideals, shouldn't the fact that $\mathrm{Aut}_{\mathrm{Ring}}(R)$ acts transitively on the maximal left ideals be reflected by the fact that $\mathrm{Aut}_{R-\mathrm{Mod}}(IR)$ acts transitively on maximal submodules?
I'm answering on the basis that the question is actually
Yes, in this particular ring, ring automorphisms transitively map maximal left ideals to maximal left ideals. And, in this particular ring, all automorphisms are inner, so this is all done by conjugation.
So given two maximal left ideals $L_1$ and $L_2$, $L_2=X^{-1}L_1X$ for some invertible matrix $X$. Since $L_1$ is a left ideal, $L_2=L_1X$.
$X$, operating this way on the right of elements of $R$, is clearly an isomorphism of left $R$ modules between $L_1$ and $L_2$ with inverse given by right multiplication by $X^{-1}$.
That shows the left $R$ linear isomorphisms of $R$ act transitively on maximal left ideals.
So the ring isomorphisms are related to the module isomorphisms, just apparently not in the way you expected.