Is any finite subset of $S^2$ satisfying the following properties, the vertex set of a regular convex polyhedron?

Question

Let $S^2 \subset \mathbb{R}^3$ denote the standard unit sphere.

Let $A \subset S^2$ be a finite set satisfying each of the following properties:

1. For each $a,a' \in A$ there exists a rotation $R$ (the action of a 3x3 orthogonal matrix) such that $R(A) = A$ and $R(a) = a'$.
2. The set $A$ is not contained in any one plane. (Such a plane need not pass through the origin.)

My question is, in the above situation, is $A$ necessarily the set of vertices of some regular convex polyhedron (embedded in $\mathbb{R}^3$)?

Answer 1

Look at the Archimedean solids for which

A global isometry of the entire solid takes one vertex to the other while laying the solid directly on its initial position.

There may be other examples.