The strong orbit-equivalence rigidity conjecture

Let $G$ and $H$ be closed permutation groups on a countably infinite set $X$. They are strongly orbit-equivalent when they have the same orbits on the power set

$$\mathcal P(X).$$

Strong orbit-equivalence rigidity conjecture. If $G$ and $H$ are strongly orbit-equivalent, then

$$H=G.$$

This strengthens ordinary orbit-equivalence by requiring agreement on all subsets, not only finite subsets. The closedness hypothesis is necessary: the source cites a proper nonclosed subgroup of the infinite symmetric group that is strongly orbit-equivalent to it. The conjecture itself remains open in the source.