I am currently reading Stan Wagon's Banach-Tarski Paradox book, and this was left as an exercise to prove (converse of Proposition 1.10).
Let $X$ be a set, and let $G$ act on $X$ with no nontrivial fixed points. If $X$ is $G$-paradoxical, then $G$ is $G$-paradoxical.
A hint was given with this, that I should transfer the paradoxical decomposition of a single orbit to $G$.
Here's how I've gone about proving this.
Proof. Let $A_{i},B_{j}\subseteq X$ and $g_{i},h_{j}\in G$ realize that $X$ is $G$-paradoxial (that is, $\left\{A_{i}\right\}\cup\left\{B_{j}\right\}$ is a partition of $X$ with $\bigcup_{1\leq i\leq m}g_{i}A_{i}=X=\bigcup_{1\leq j\leq n}h_{j}B_{j}$). Now, let $x\in X$. Then, the family of subsets consisting of $A_{i}^{*}=\left\{g\in G\mid gx\in A_{i}\right\}$ and $B_{j}^{*}=\left\{g\in G\mid gx\in B_{j}\right\}$ is a partition of $G$, as there are no nontrivial fixed points in the $G$-action on $X$, such that $\bigcup_{1\leq i\leq m}g_{i}A_{i}^{*}=G=\bigcup_{1\leq i\leq n}h_{j}B_{j}^{*}$. Therefore, $G$ is $G$-paradoxical. QED
My reasoning for the last equality: Since $\bigcup_{1\leq i\leq m}g_{i}A_{i}=X=\bigcup_{1\leq j\leq n}h_{j}B_{j}$, we would have $\bigcup_{1\leq i\leq m}g_{i}A_{i}^{*}\mathcal{O}_{x}=\mathcal{O}_{x}=\bigcup_{1\leq i\leq n}h_{j}B_{j}^{*}\mathcal{O}_{x}$ and, since the $G$-action has no nontrivial fixed point, there's a bijective correspondence $\mathcal{O}_{x}\leftrightarrow G$ given by $g\cdot x\leftrightarrow g$.
I feel like this is how one would prove this, but I need to know if this is correct. In particular, I feel the most unsure about the last equality statement, which is why I listed the reasoning above. Is this all correct?
EDIT:
Here's the final draft of my proof.
Proof. Let $A_{i},B_{j}\subseteq X$ and $g_{i},h_{j}\in G$ realize that $X$ is $G$-paradoxial (that is, $\left\{A_{i}\right\}\cup\left\{B_{j}\right\}$ is a partition of $X$ with $\bigcup_{1\leq i\leq m}g_{i}A_{i}=X=\bigcup_{1\leq j\leq n}h_{j}B_{j}$). Now, let $x\in X$. Then, the family of subsets consisting of $A_{i}^{*}=\left\{g\in G\mid gx\in A_{i}\right\}$ and $B_{j}^{*}=\left\{g\in G\mid gx\in B_{j}\right\}$ is a partition of $G$, as there are no nontrivial fixed points in the $G$-action on $X$. Since $\bigcup_{1\leq i\leq m}g_{i}A_{i}=X=\bigcup_{1\leq j\leq n}h_{j}B_{j}$, we have $\bigcup_{1\leq i\leq m}g_{i}A_{i}^{*}\mathcal{O}_{x}=\mathcal{O}_{x}=\bigcup_{1\leq i\leq n}h_{j}B_{j}^{*}\mathcal{O}_{x}$ and there's a bijective correspondence $\mathcal{O}_{x}\leftrightarrow G$ given by $g\cdot x\leftrightarrow g$. So, stripping away the $x$ using this bijection, we have that $\bigcup_{1\leq i\leq m}g_{i}A_{i}^{*}=G=\bigcup_{1\leq i\leq n}h_{j}B_{j}^{*}$. Therefore, $G$ is $G$-paradoxical. QED
I think it is simpler. Take any x and let A be its orbit. THe action of G works when completely restricted to A. Therefore the paradoxical decomp. must induce two subsets of A that each give the full copy of A. That is, we can simply restrict everything in sight to A. Then, because of the lack of fixed points, this induces what you want on G. A is simply a copy of G. A looks like {x, g1 x, g2 x, g3 x,...} where the g's enumerate g.
This is just a sketch, but it appears to be different than what you were doing.
SW