Given a polyhedron $P$ with vertex set $V$ of size $n$, say that it is subset-symmetric if, for every subset $S\subseteq V$, there is a non-identity element of the full reflectional symmetry group of $P$ which fixes $S$ (not necessarily pointwise). Obviously, it suffices to check this property for $|S|\le\frac n2$.
One can verify by hand that the cube has this property, which is larger/more complicated than I would have intuitively guessed.
Can it be proven that the cube has the most vertices of any subset-symmetric polyhedron?
I would also be interested in a complete classification of such polyhedra. The ones I am aware of:
Any pyramid or bipyramid whose base is an equilateral triangle, a rectangle, or a regular pentagon
The cube
(One could also ask this about symmetries of subsets of the edges of a polyhedron, in which case the cube fails, and I am not sure if any polyhedron besides the regular tetrahedron has this property.)