Let $(R, +, \cdot)$ be an associative commutative nilpotent ring of cardinality $2^n$ such that $$ r^2 = 0, $$ for every $r\in R$. Also $(V, +)$ is a vector space over $\mathbb{F}_2$. Let $\operatorname{Aut}(R)$ be a group of ring automorphisms. That is a group of bijective maps $\phi: R\to R $ such taht for every $r,s\in R$ and $*\in \{+, \cdot\}$ $$ \phi(r*s) = \phi(r)*\phi(s). $$
I want to know how the group of automorphisms for the direct product $$ \underbrace{R\oplus R \oplus\ldots \oplus R}_{k} $$ is structured.
Of course it contains trivial diagonal automorphisms of the form $$ \Phi(r_1, \ldots, r_k) = (\phi_1(r_1), \ldots, \phi_k(r_k)), $$ where $\phi_i\in \operatorname{Aut}(R)$. Also there are automorphisms acting as permutations, that is for $\sigma \in S_k$ $$ \Psi_{\sigma}(r_1, \ldots, r_k) = (r_{\sigma(1)}, \ldots, r_{\sigma(k)}). $$
But what about more subtle maps? It is interesting how the whole group of automorphisms for $\underbrace{R\oplus R \oplus\ldots \oplus R}_{k}$ is structured.
My attempts. I'm trying to start from investigation for $$ R\oplus R $$ automorphisms. Let $\phi$ be such automorphism. Then since $\phi$ is $+$ automorphism there exist $\alpha, \beta, \gamma, \delta \in \operatorname{Hom_+}(R,R)$ such that $$ \phi(x,y) = (\alpha(x) + \beta(y), \gamma(x) + \delta(y)), $$ (this is usual matrix product). Then since $\phi$ preserves multiplication we have $$ \phi(xz,yt) = \phi(x,y)\phi(z,t), $$ for arbitrary $x,y,z,t\in R$. Combining this with the above equality and using the fact that we working in vector space over $\mathbb{F}_2$ we can obtain that $\alpha,\beta, \gamma, \delta$ are also multiplicative homomorphisms. Then $\phi$ is automorphism if and only if for all $x,y,z,t\in R$ we have $$ \begin{array}{l} \alpha(x)\beta(t) = \beta(y)\alpha(z) \\ \gamma(x)\delta(t) = \delta(y)\gamma(z) \\ \end{array} $$ and $\phi$ is bijection. Next, I'm stuck.