Let $R$ be a commutative ring with unity and $\nu: R^n \rightarrow R^n$ is a nilpotent endomorphism ($R$- module homomorphism). If no other conditions are given on $\nu$ apart from being nilpotent then is it possible to construct two non trivial (apart from id,-id) $R$ module automorphisms from $R^n \rightarrow R^n$ that commutes with $\nu$.
What I would like to know is does there exist two automorphisms, $f \neq g$ and $f,g: R^n \rightarrow R^n$ such that $\nu f = f \nu$ and $\nu g = g\nu$. I know the question itself does not pose a good look, but I was curious if at all I could construct two automorphisms like this apart from id and -id.
Let us take some polynomial of form $P(x) = a_d x^d + a_{d-1} x^{d-1} + \ldots + a_2x^2 + a_1x + 1$ with $a_i \in R$. Then I claim that $P(\nu)$ is an automorphism of $R^n$, and that $\nu P(\nu) = P(\nu) \nu$.
The last equation is clear: we get $a_d \nu^{d+1} + a_{d-1} \nu^d + \ldots + a_1 \nu^2 + \nu$ both ways, $R$ is commutative so everything works fine. For the first part, let $m$ be such that $\nu^m=0$. Then we will try to find $Q(x)$ such that $Q(\nu)P(\nu)$ is identity. Let $Q(x) = 1+b_1x + b_2x^2 + \ldots + b_{m-1}x^{m-1}$. We want $P(x)Q(x)$ to have coefficients by $x^k$ equal to $0$, for $k=1,2,\ldots,m-1$. This gives us some equations for $b_i$: $a_1+b_1=0$, $b_2+a_1b_1+a_2=0$, and so on. Using these equations, we may recursively determine $b_i$-s and find appropriate $Q(x)$.
This way you can find at least $|R|^{m-1}$ automorphisms with your desired property.