I am proving the Banach-Tarski paradox using a series of small results. For definition of certain terms, see here.
Group $G$ acts freely on $X$ i.e. $\operatorname{Stab}(x)=e, \ \forall \ x\in X$. Suppose $G$ is paradoxical (i.e. $G$ is equidecomposable with 2 copies of $G$) w.r.t. left translation action of group $G$. Then $X$ is equidecomposable with 2 copies of $G$.
I was given the hint that axiom of choice is required.
Progress: By AC, from each distinct orbit we choose an element so that together they form a set say $S$. We define a map $\phi : G\times S \to X$ as $\phi(g,s)=gs$. We define a group action on $G\times S$ as $h*(g,s)=(hg,s)$. Then $\phi(g_2*(g_1,s))=g_2\phi(g_1,s)$. With respect to this new group action $G\times S$ is paradoxical. Then can we claim $X$ is paradoxical? What is the justification here?