# Is there a $G$-paradoxical $G$-invariant subset of the plane for $G$ a group of rigid motions?

The Sierpinski-Mazurkiewicz paradox yields a nonempty rigid-motion paradoxical subset $$S$$ of the Euclidean plane: $$S$$ is the disjoint union of $$A$$ and $$B$$, each of which is $$G$$-equidecomposable with $$S$$, for a group $$G$$ of rigid motions.

(Here sets $$U$$ and $$V$$ are $$G$$-equidecomposable for a group of $$G$$ acting on a space containing $$U$$ and $$V$$ iff $$U$$ can be written as the disjoint union of $$U_1,...,U_n$$, $$V$$ as the disjoint union of $$V_1,...,V_n$$ and $$V_i = g_iU_i$$ for some sequence $$g_1,...,g_n\in G$$.)

Question: Can one do this with $$S$$ itself being $$G$$-invariant? I.e., is there is a group $$G$$ of rigid motions (variant: isometries) and a subset $$S$$ of the plane such that $$S$$ is $$G$$-paradoxical and $$gS=S$$ for all $$g\in G$$?

No, it's not possible because the isometry group of the Euclidean plane is amenable (as discrete group, since it's solvable), so every $$G$$-set admits an invariant mean defined on all subsets. If $$S$$ is $$G$$-invariant, this applies to $$S$$ as $$G$$-set and an immediate contradiction follows.