Consider a group $G$ acting on a set $X$. The action of $G$ is free if and only if $g(x)=x\Rightarrow g=e$. Consider the $\mathbb Z[G]$-module $M(X)$ given by the free $\mathbb Z$-module generated by $X$, with the action of $G$ which is given, on generators, by $g\cdot x=g(x)$ in the original sense.
I wonder if the equivalence $G$ acts free on $X\iff M(X)$ is a free $\mathbb Z[G]$-module holds. In fact, if $G$ acts also transitively, $M(X)$ is free of rank $1$. Does the only condition "free action" guarantee that $M(X)$ is at least free as $\mathbb Z[G]$-module?
Thank you in advance.