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 $g\in G$?



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


