Is a vertex- and edge-transitive polytope already a uniform polytope? 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 a uniform 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 every Wythoffian polytopes 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).
 A: Well, your addition on the polytope being forced to be Wythoffian definitely guarantees a positive answer. In fact, every Wythoffian polytope with equal sized edges will be uniform. (Vertex transitivity is already contained when asked to be Wythoffian by means of kaleidoscopical construction.)
But not every uniform polytope is Wythoffian. Examples are eg. the snub figures and Coxeter's grand antiprism. Those cannot be constructed via mere kaleidoscopical constructions.
--- rk
A: A vertex- and edge-transitive polyhedron is already a uniform polyhedron.
This is because the vertex-transitivity asures the vertices to be equivalent and therefore the angles of each face (individually) to be the same. The edge-transitivity then asures the edges to be all of the same size (say unity). And so the faces individually to become regular ones.
But be aware, that there are uniform polyhedra, which aren't edge-transitive!
Within higher dimensional spaces, assuming spherical geometry, these axioms only provide scaliformity. That is, you well could use some of the Johnson solids to be cells of your polytope.
The easiest scaliform polychoron (4D polytope) is the lace prism of 2 truncated tetrahedra in inverted orientation. That is the lacing cells here are 8 triangular cupolae and 6 tetrahedra, while the top and bottom base are those truncated tetrahedra.
But again, scaliformity just asks the edges to be of the same size. Not that those have to be equivalent. So your axioms select a subset out of the scaliforms.
--- rk
