This is not possible in general, or even in particular when $n=3$. See this paper of Ziegler for a reference.
I would guess that this is possible if I allow the sphere to have arbitrarily high dimension (maybe you can take $S^n$ as the one point compactification of $\mathbb{R}^n$ and use stereographic projection to inscribe your $n$-polytope in $S^n$, but I'm not sure.)
I would address this question from the point of view of group theory. Namely, if one starts to classify finite subgroups $G$ of $\mathrm{SO}(3)$ (say, first looking at thier poles – points where an rotation axis intersects the sphere $S^2$), then it's not difficult to prove by orbit counting that there are following possibilities for finite subgroups of $\mathrm{SO}(3)$:
- cyclic
- dihedral
- some group $\mathbf T$ of order 12 with three types of pole orbits of sizes 4, 6 and 4 (hence with stabilisers of sizes 3, 2 and 3);
- some group $\mathbf O$ of order 24 with three types of pole orbits of sizes 8, 12 and 6 (hence with stabilisers of sizes 3, 2 and 4);
- some group $\mathbf I$ of order 60 with three types of pole orbits of sizes 12, 30 and 20 (hence with stabilisers of sizes 5, 2 and 3).
Of course, the tricky point is to check that $\mathbf T$, $\mathbf O$ and $\mathbf I$ exist without appealing to the existence of Platonic solids :) This can be done in at least two ways:
- using explicit presentations of them as described, for instance, in § 57 of the book G. A. Miller, H. F. Blichfeldt, L. E. Dickson, Theory and applications of finite groups, Dover, New York, 1916; for instance, for the tetrahedron it reads $s_1^3=s_2^3 = (s_1s_2)^2 = 1$ which is easily realisable by rotations;
- or (amazingly!) using Riemann surfaces.
Now, if we take any orbit whose stabiliser has size more than 2, (meaning an orbit of size 4 for $\mathbf T$, an orbit of size 8 or 6 for $\mathbf O$, an orbit of size 12 or 20 for $\mathbf I$), then its points will define vertices of a regular polytope (whose edges can be defined as connecting a vertex with nearest vertices).
Each vertex can be rotated by $G$ to another one by construction, and the stabiliser of each vertex consists of rotations around it, which clearly have to permute edges going out of this vertex. Now, the sizes of stabilisers guarantee that there are exactly as many edges going out of each vertex as the order of the stabiliser, and therefore every edge can be rotated to every other edge by an element of $G$. As $G$ clearly preserves faces of our polyhedron, they are forced to be regular polygons. An inspection of orders of stabilisers thus gives the list of Platonic solids.
Notice that the “exceptional” orbits have stabilisers of order 2, so the above construction doesn't work there because there's not enough rotations around these vertices (however, we were only concerned with existence of Platonic solids anyway).
Best Answer
I think the intuition here is 'triangles are better than squares'. Squares aren't always avoidable, but a triangular configuration of vertices is more 'rigid' than a square one; it's easier to make small perturbations of the verts of a square that increase all the distances. In this case, by twisting the upper face $45^\circ$ you're increasing the minimum distance between verts on the upper face and those on the lower face, which means that you can in turn 'compress' those two faces closer to each other (i.e., bring them closer towards the equator of the sphere) to make the side lengths of the faces bigger while still maintaining inscribability. Similarly, while the icosahedron is optimal for 12 points because of the triangular facets, I would be very surprised if the best 20-point configuration were dodecahedral.