If $M\otimes M^\vee\rightarrow \operatorname{End}(M)$ is an isomorphism then $M$ is reflexive (hence dualizable). 
Let $A$ be a (commutative if necessary) ring and $M$ an $A$-module. Suppose that the map
\begin{align}M\otimes M^\vee&\longrightarrow \operatorname{End}(M) \\ m\otimes \phi(\cdot)&\longmapsto m\phi(\cdot)\end{align} is an isomorphism. Is it true that then the canonical map $M\longrightarrow (M^\vee)^\vee$ has to be an isomorphism?

Here $M^\vee=\operatorname{Hom}(M,A).$
Context:
In a monoidal category there is a concept of dualizable object. Following Duality, Trace and Transfer by Dold, Puppe, if one takes the category to be $A$-Mod, a corresponding dualizable module is a reflexive module (that is, $M\overset{\sim}{\rightarrow} (M^\vee)^\vee$) in which $M\otimes M^\vee$ is canonically self dual. But if $M$ is reflexive then $$\operatorname{End}(M)=\operatorname{Hom}(M,M)=\operatorname{Hom}(M,\operatorname{Hom}(M^\vee,A))=\operatorname{Hom}(M\otimes M^\vee,A)=(M\otimes M^\vee)^\vee$$
Hence, canonically self dual is equivalent to $M\otimes M^\vee\cong \operatorname{End}(M)$ under reflexivity. Hence, if we have the implication above, a dualizable module would be exactly the modules in which $M\otimes M^\vee\cong \operatorname{End}(M)$ is an isomorphism, which I think would be very nice.
A remark, there is a characterization of dualizable modules, they are exactly the finitely generated projective modules over $A$ (Here $A$ is not necesarilly commutative). Hence, a possible counterexample should be either non projective or non finitely generated.
 A: I manage to do it using the ideas in the proof of the reference in this answer found by Vladimir Sotirov. Here it is the final conclusion:
Theorem. For a ring $A$ and a module $M$ the following are equivalent:

*

*$X \otimes M^\vee\rightarrow \operatorname{Hom}(M,X)$ is an isomorphism for any module $X$.

*$M\otimes M^\vee\rightarrow \operatorname{End}(M)$ is surjective

*$M$ is finitely generated and projective

*$M$ is a dualizable object in $(A$-Mod$,\otimes)$.

Under these condictions $M$ is a dualizable module. In particular, $M$ is reflexive.
Proof:
1 implies 2: Take $X=M$
2 implies 3: By surjectivity there are $\{m_i\}_i, \{\varphi_i\}_i$ such that $1_M=\sum_{i=1}^n m_i\otimes \varphi_i$. Then as
$$m=1_M(m)=\sum_{i=1}^n m_i\otimes \varphi_i(m) \quad \forall m\in M,$$
$M$ is generated by $\{m_i\}$ and the pair $(\{m_i\}_i, \{\varphi_i\}_i)$ is a projective basis, so the module is finitely generated and projective.
3 implies 1: The map is an isomorphism when $M=A^n$. Moreover, if it is an isomorphism for $A^n=N\oplus N'$, then it is an isomorphism for each direct summand $N,N'$.
The fact that $M$ is a dualizable object in this case was the comment in the question, and the fact that $M$ reflexive is similar to 2 implies 3 above.
