Let $M$ be a simple module over a unital ring $R$. If $|M| \leq 2$, then $M$ has only one automorphism (the identity). I'm wondering whether for $|M| > 2$, there always exists an automorphism distinct from identity. I showed that this is true for commutative $R$. In fact in this case for any nonzero elements $x,y \in M$, there is an automorphism that maps $x$ to $y$.
In the case of noncommutative $R$, I wasn't able to prove the existence of a nontrivial automorphism or find a counterexample. I'm mostly interested in the case when $M$ and $R$ are finite. I showed that for any counterexample one would have that $|R| \geq |M|^2$. However I'm too inexperienced with modules to find a counterexample if there is one.
If this is in fact also true for noncommutative $R$, I would appreciate a reference to this result.
It turns out it is not always the case for simple modules over noncommutative rings. One counterexample is the simple module $\mathbb{Z}_2 \times \mathbb{Z}_2 = \Big\{\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big\}$ over the ring of $2 \times 2$ matrices over $\mathbb{Z}_2$. Let $A = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$. Since $A\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ and both $A\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ and $A\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are distinct from $\begin{bmatrix} 0 \\ 0 \end{bmatrix}$, there is no automorphism that maps $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to either $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ or $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Thus an automorphism can only map $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to itself. However an automorphism of a simple module is determined by its value on a single nonzero argument, so it follows that the only automorphism is the identity.
Note that this counterexample obeys the restriction that I mentioned in my question. In fact if $|R| < |M|^2$, then for any nonzero elements $x,y \in M$, there is an automorphism that maps $x$ to $y$, just as in the commutative case.