Given a commutative ring $A$ and an $A$-module $M$, does there exist $A$-modules $M'$ and $M''$ so that $M\cong\operatorname{Hom}_A(M',M'')$?
So far, I have that if $M$ is free of finite rank then there does exists $A$-modules $M'$ and $M''$ so that $M=\operatorname{Hom}(M',M'')$. To see this, let $M$ be free of finite rank, so
$$M\cong M^{\vee \vee}=\operatorname{Hom}_A(M^{\vee},A)$$
(where $M^{\vee}=\operatorname{Hom}_A(M,A)$). Since $A$ is commutative, $M^{\vee}$ is a module over $A$, so $M$ is of the desired form.
I cannot make progress in the more general question, so any help will be appreciated. Thanks!
How about $M\cong\mathrm{Hom}_A(A,M)$ with the natural identification $m\mapsto(a\mapsto am)$?