Let $R$ be a commutative Noetherian ring. For any $R$-module $M,N$, there exists a canonical map $d_{M,N}: M \to \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$ sending an element $m \in M$ to the map $\operatorname{Hom}_R(M,N) \to N$ $(f \mapsto f(m))$. Now let $M$ be a finitely generated reflexive $R$-module. If $N \cong R$ , then $M\cong \operatorname{Hom}_R(\operatorname{Hom}_R(M,N),N)$.
My question is: If $M$ is a finitely generated reflexive $R$-module and $N\cong R$, then is the map $d_{M,N}$ an isomorphism?
Let $\varphi:R\to N$ be an isomorphism, then it induces an isomorphism $$\tilde\varphi:\hom_R(\hom_R(M,N),\,N)\to\hom_R(\hom_R(M,R),\,R)\\ \tilde\varphi(F)=\underset{M\to R}f\mapsto \varphi^{-1}\left( F(\varphi\circ f)\right)\,$$ and for the evaluation map $d_{M,N}$, i.e. setting $F=d_{M,N}(m)$ in the above, we get $$\tilde\varphi\circ d_{M,N}(m)=\underset{M\to R}f \mapsto\varphi^{-1}(\varphi\circ f(m))=f\mapsto f(m)=d_{M,R}(m)\,.$$ So, $d_{M,N}$ is an isomorphism if and only if $d_{M,R}$ is.