I just learned the existence of Gil Kalai's $3^{d}$ conjecture, which according to Wikipedia, is proven for $d$ at most $4$. It states that every $d$ dimensional polytope with central symmetry has at least $3^{d}$ "faces" (vertices, edges, true faces and so on, including the polytope itself but not the empty set). As Wikipedia gives the example of the parallelogram and the cube and itsdual the octahedron, I came to think that maybe there could be a combinatorial interpretation of this conjecture in terms of the symmetry group of an "orthogonalized" version of the polytope or of its dual, that is a $d$-dimensional parallelotope, which would have at least $(\mathbb{Z}/2)^{d}$ as isometry group and which would thus entail the central symmetry property.

The number $3^{d}$ would then emerge from taking $0$, $1$ or $2$ elements of each copy of $(\mathbb{Z}/2)$ and combining them to get a subgroup of the isometry group of a "face". It would then remain to show that any given "flexible" polytope or the dual thereof can be continuously deformed into such a parallelotope (necessarily of same $d$-dimensional volume, as Lebesgue measure) with corners cut out by hyperplanes in such a way that the central symmetry is preserved (a repetition of such "CS cuts" being possibly needed to recover the considered polytope).

Has such an approach been considered so far to tackle this problem? If so, can you give some reference?

Of course, Gil himself is more than welcome to express his feeling about these ideas as well as his original motivation in formulating this conjecture.

Edit: it seems the notion of "vertex figure" as defined in https://en.m.wikipedia.org/wiki/Vertex_figure corresponds to the idea of CS cut I had in mind, but I'm not sure it helps much. Maybe we should first classify uniform polytopes one can obtain that way from hypercubes.