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


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



Dual of a semilinear morphism
No.


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 $y$.


