Start with the Hilbert cube $H=I^\omega$, thinking of its coordinates as written in ternary expansion.
Construct subsets $S_n$ by removing points from $H$ if for any $m$, at least $n$ of the coordinates must have a ternary digit equal $1$ in the $m^{\rm th}$ place. So $S_0$ equals the empty set, $S_1$ equals a countable product of standard Cantor sets (and hence a Cantor set unto itself); $S_2$ projected onto any two coordinates gives a traditional Sierpinski carpet, etc.
If I understand correctly, the natural actions of the autohomeomorphism group of the standard Sierpinski carpet has just two orbits and the natural action of the autohomeomorphism group of the Hilbert cube (surprisingly) has just one orbit.
So how many orbits for $S_n, n>1$, under natural actions by their autohomeomorphism groups?
