This question is from 'Cohomology of Groups' by Atiyah and Wall, p.95 of Cassels' and Frohlich's book 'Algebraic Number Theory'.
Let $G$ be a group and $A={\rm Hom}_{\mathbb Z}(\mathbb Z[G],X)$ where $X$ is an abelian group. $A$ is a $G$-module via $(g\phi)(a)=(g(\phi(g^{-1}a))$ for all $g\in G$, $\phi \in A$, $a\in \mathbb Z[G]$. Let $B$ be any $G$-module. The book says that there is an isomorphism $${\rm Hom}_G(B,A)\cong {\rm Hom}(B,X)$$ of $G$-modules given by $\phi\mapsto (b\mapsto \phi(b)(e))$ where $e$ is the identity of $G$.
Firstly, is the $G$-action on $X$ the trivial one? Assuming this, I can see that it is a $G$-homomorphism but why is it a bijection? What is the inverse of this map? Is it given by $f\mapsto (b\mapsto (g\mapsto f(b)))$ for all $g\in G$?
Given $f \in {\rm Hom}(B,X)$, the element $\phi$ mapping onto $f$ has to be a $G$-homomorphism from $B$ to ${\rm Hom}({\mathbb Z}G,X)$.
So, for $g \in G$, $b \in B$, we must have $\phi(gb) = g\phi(b)$. So, $f(gb) = \phi(gb)(1) = (g\phi(b))(1) = g(\phi(b)(g^{-1})$, and hence $\phi(b)(g^{-1}) = g^{-1}f(gb)$.
So the inverse map you are looking for is $f \mapsto (b \mapsto (g \mapsto gf(g^{-1}b))$.