$G$ acts $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$

Question

Let $A$ be a $G$-module where the action $G$ on $A$ is trivial. Then also $G$ acts on $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$ such that $g\cdot f(x):=f(xg)$. On the other hand, we know that $\mathbb{Z}G$ is a free $\mathbb Z$-module with rank $|G|$ and so we can conclude that $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$ is homomorphic to $A^{|G|}$ as abelian groups. Since $G$ acts on $\rm{Hom}_{\,\mathbb Z}(\mathbb{Z}G,A)$, we must have an action $A^{|G|}$. But I could not find how $G$ acts on $A^{|G|}$.

Answer 1

Elements of $G$ act as permutations on $G$ by $h\mapsto hg.$ The action of $G$ on $A^{|G|}$ (i.e., on the full set of maps from $G$ to $A$) is defined by composition with these permutations.