It's easy to prove that the convex hull of any vertex-transitive polyhedron is vertex-transitive. Specifically, any symmetry of the original polyhedron that moves any vertex to another will also move the same vertices on the convex hull, but since it will preserve the general shape of the original polyhedron (and therefore the set of vertices) it must preserve the convex hull.
However, my question is the following: Are the convex hulls of edge-transitive polyhedra edge-transitive, and are the convex hulls of face-transitive polyhedra face-transitive?
I have gathered evidence backing up this, mostly from uniform polyhedra, but I can't use the same technique as before to support my claim. Does anyone know why is it true or does anyone have a counterexample?
I'm going to assume that we're dealing with polyhedra whose faces are planar polygons (possibly self-intersecting, like a pentagram, but contained in a plane), with no two co-planar.
Here's a very sketchy proof sketch.
Now, in Grünbaum and Shephard's article Duality of Polyhedra, they are attempting to grind the axe "Duals are not as well-defined as everyone seems to think they are", but what they actually say is that for polyhedra with planar, possibly self-intersecting faces, a dual can always be constructed by polarity with matching symmetry/transitivity/regularity properties.
In particular, a face-transitive polyhedron has all its face centroids lying on a sphere, the center of which is a fixed point of its symmetry group, which is the natural choice for the center of the sphere of reciprocation.
So, a face-transitive polyhedron $P$ has a dual $P^*$ which is vertex-transitive. As you say, $\operatorname{conv}(P^*)$ is a vertex-transitive convex polytope, and so its dual, $\operatorname{conv}(P^*)^*$ is a face-transitive convex polytope.
Now, the question is, is the dual of the convex hull of the dual the same as the convex hull of the original polytope? I.E. does $\operatorname{conv}(P^*)^* = \operatorname{conv}(P)$? And my claim is "Sure, what else would it be?"
That leaves the case of edge-transitive polyhedra. But every edge-transitive polyhedron is also either vertex-transitive or face-transitive.
You can find this claim on the Wikipedia page Isotoxal figure ("An isotoxal polyhedron or tiling must be either isogonal (vertex-transitive) or isohedral (face-transitive) or both."), and also in Peter Cromwell's 1999 book Polyhedra, in both cases appearing to apply to star polyhedra as well as convex ones. I'm sure that I once dug up a proof of this, but I forget where.
Perhaps you can make some progress closing up the gaps I left you. ;-)