I am looking for an example of a ring $R$, a left $R$-module ${}_{R}{M}$ and a ring automorphism $\psi:R \rightarrow R$ such that the derived module ${}_RM^\psi$ is not isomorphic to ${}_RM$. Note: I am only looking at morphisms for which $\psi(1)=1$.
${}_RM^\psi$ is defined as having the same abelian group and the exterior product $r*m:= \psi(r) \cdot m$, where $\cdot$ is the product in ${}_RM$.
The main difficulty I'm having is broader, namely that I barely know any "simple" rings which have nontrivial automorphism groups; there's $\mathbb{C}$, but its wild automorphisms are not given explicitly and seem therefore hard to work with, if not impossible. Perhaps polynomial rings might have concrete non-trivial automorphisms?
Thank you for your time and any suggestions.
Let $R=\mathbb{F}_2^2$ and let $M=\mathbb{F}_2$. Then $M$ is an $R$-module via $(\bar{m},\bar{n})\cdot \bar{x}=\bar{m}\bar{x}$ for all $\bar{m},\bar{n},\bar{x}\in\mathbb{F}_2$.
Let $\psi$ be the ring automorphism of $R$ switching coordinates. Assume now that $u:M\to M^\psi$ is an isomorphism of $R$-modules. In particular, it is an automorphism of the additive group $\mathbb{F}_2$. Hence $u=\mathrm{Id}_{\mathbb{F}_2}$. Now for all $\bar{m},\bar{n},\bar{x}\in\mathbb{F}_2$, we should have $\psi(\bar{m},\bar{n})\cdot u(\bar{x})=\bar{n}\bar{x}=u((\bar{m},\bar{n})\cdot\bar{x})=\bar{m}\bar{x}$, which is not the case.
Consequently, $M$ and $M^\psi$ are not isomorphic as $R$-modules.