I want to consider (convex) polytopes $P=\mathrm{conv}\{p_1,...,p_n\}\subset\Bbb R^d$ which are both, *vertex- and edge-transitive* (or maybe stronger: *1-flag-transitive*).

Question:Is every such polytope already auniform polytope?

I know only a few polytopes with such symmetries, all of them are uniform, probably also Wythoffian. Here are some:

- Regular polytopes,
- Vertex truncations (rectification) of some regular polytopes like the $d$-simplex or the $d$-cube,
- Hypersimplices,
- Exceptional polytopes like the $2_{21}$-polytope or $3_{21}$-polytope,
- Cartesian product of two or more identical copies of one of the above (e.g. duoprisms),
- ...

The faces of uniform polytopes are uniform again. So far, all I can say about the faces of vertex- and edge-transitive polytopes is, that they are polytopes with all vertices on a sphere and all edges of the same length. While this means that all 2-faces are uniform, it does not immediately follow for the 3-faces (e.g. Pseudorhombicuboctahedron is not uniform but *could* be a face).

I know that after all, the Wythoffian uniform polytopes are the most well understood. Also, I do not know whethe there is any non-Wythoffian vertex- and edge-transitive polytope. So as a first step, I might ask:

Question:Is everyWythoffianpolytopes with such symmetries already uniform?

**Update**

As mentioned by Dr. Klitzing, the second question seems to be trivial, as a Wythoffian polytope is always uniform as long as its edge lengths are equal everywhere (which surely is the case for edge-transitive polytopes).