# Dual of a semilinear morphism

Let $R$ be a commutative ring and let $M$ and $N$ be $R$-modules. Let $\sigma:R\rightarrow R$ be a ring automorphism.

Let $f: M\rightarrow N$ be a $\sigma$-semilinear map, i.e. a map of abelian groups satisfying $f(rm)=\sigma(r)f(m)$. Let $Hom_{R,\sigma}(M,N)$ denote the set of all $\sigma$-semilinear maps from $M$ to $N$.

The usual definition of the dual of $f$ does not give the right result, as it is a map $N^*=Hom_{R,id}(N,R)\rightarrow Hom_{R,\sigma}(M,R)\qquad n^* \mapsto n^* \circ f$. To force the target to be $Hom_{R,id}(M,R)$ I would like to postcompose with $\sigma^{-1}\in Hom_{R,\sigma^{-1}}(R,R)$. But of course the $\sigma$ is not uniquely determined by the map $f$. For example the zero map is $\sigma$-semilinear for any $\sigma$.

So my question is: Is this well defined, i.e. Given any $f\in Hom_{R,\sigma}(M,N)\cap Hom_{R,\sigma'}(M,N)$ and any $n^* \in N^*$. Do we have

$\sigma^{-1} \circ n^* \circ f= \sigma '^{-1}\circ n^* \circ f$?

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Take $R=k[x,y]/(x^2,xy,y^2)$ and let $\sigma=id$ and $\sigma'$ swap $x$ and $y$.
Let $M=N=R/(x,y)$, then the identity map $M\to N$ is $\sigma$ and $\sigma'$ semilinear.
One element of $N^*$ sends the generator to $x$. But $\sigma^{-1}$ sends $x$ to itself and $\sigma'^{-1}$ sends $x$ to $y$.