$\DeclareMathOperator\Hom{Hom}$Let $M$ be a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$. Denote $(\_)^*:=\Hom_R(\_,R)$. There is a natural map \begin{align} \Phi_M: M \otimes_R M^* & \longrightarrow \Hom_R(M^*,M^*) \\ x \otimes y & \longmapsto (f\mapsto f(x)y). \end{align}

My question is:

If there exists an isomorphism $M\otimes_RM^*\cong \Hom_R(M^*,M^*)$, then is it true that $\Phi_M$ is an isomorphism?

My reason for asking this question:

It is well known that if $M$ is a finitely generated module over a Noetherian local ring $(R,\mathfrak m)$ such that $M\cong M^{**}$, then the natural evaluation map \begin{align} M & \longrightarrow M^{**} \\ x & \longmapsto (f \mapsto f(x)) \end{align} is an isomorphism (see the book Gorenstein dimensions, Proposition (1.1.9)). The map defined in my question is kind of similar, so I wonder if that is true as well.

My thoughts:

If we have isomorphic finitely generated modules $X \cong Y$ and a linear surjection $f:X\to Y$, then $f$ is an isomorphism. Indeed, let $g:Y\to X$ be an isomorphism. Then, $g\circ f: X\to X$ is a surjection, hence it is an isomorphism (by Nakayama), so $f$ is injective, and we are done. Hence, under the hypothesis of my question, it would be enough to show $\Phi_M$ is surjective….