Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid? Suppose that we have a convex polyhedron $P$, such that the symmetry group of $P$ acts transitively on its vertices, edges, and faces (that is, it is isogonal, isotoxal, and isohedral). It then follows that $P$ is one of the Platonic solids, which have a number of equivalent definitions but usually includes some stipulation that the faces are regular. (Once you have regular faces plus these symmetry conditions, most other properties follow easily.)
This can be shown via a rather gross casework argument on the number of sides at each face. By face-transitivity, all faces are congruent (in fact, we only need that the polyhedron be monohedral for this). By edge-transitivity, said faces are equilateral polygons. Then we break into cases by the number of sides of the polygons:

*

*If the faces are triangles, being equilateral implies being regular.


*If the faces are non-regular equilateral quadrilaterals, they are rhombi. Then, at every vertex there must be an equal number of "skinny" and "wide" angles meeting, and there must be at least 3 faces meeting at a vertex. But then there are at least two skinny and two wide angles (since we assume the angles are different), so the angles at each vertex add up to at least 360 degrees, which is not possible in a convex polyhedron.


*If the faces are non-regular equilateral pentagons, then they have angles which are not all the same; however, each vertex must have the same angles around it, so each vertex has a proportional fraction of the angles of the pentagons. But for any distribution other than the uniform one, the only way to have a proportional amount of every angle is to have some $n$ copies of every angle with multiplicity, since $5$ is prime. (E.g. if there are $2$ angles of one size and $3$ of another, we can't get a $2:3$ ratio without having at least $5$ angles at every vertex.) But the angles of a pentagon add to more than $360^\circ$, so we can't have them all at once.


*If the faces are hexagons or higher, they cannot form a topological sphere by Euler's formula.
However, I don't like this proof very much; I walk away from it with no understanding of why this behavior occurs, and it certainly doesn't suggest any generalizations to other dimensions (where I believe the analogous statement - that transitivity on every degree of cell in a convex polytope implies regularity - is true, though I would appreciate a reference for this fact).
Is there a natural or more elegant proof of this statement?
One thing to note is that any proof must somehow use the fact that we are working on the sphere. In the case of planar tilings, the corresponding implication is false: being transitive on faces, vertices, and edges does not imply that one is looking at a tiling of regular polygons! As a counterexample, consider a rhombic tiling like this one:
                                  
As another aside, relaxing any one of these conditions allows for polyhedra whose faces are not congruent regular polygons: the rhombic dodecahedron is face-transitive and edge-transitive, an isosceles tetrahedron is face-transitive and vertex-transitive, and the cuboctahedron is edge-transitive and vertex-transitive.
 A: Let a vertex have degree $\rho$ and a face have $s$ edges and note that $$\rho V=2E=sF\tag 1$$
Let $G$ be the group of symmetries. Let $G_v,G_e\text { and }G_f$ be the stabiliser of a vertex $v$, edge $e$ and face $f$, respectively. Then $$|G|=|G_v||V|=|G_e||E|=|G_f||F|\tag 2$$
If $G_e$ has a non-identity symmetry then it is a rotation of $180^o$ about the mid-point of the edge.  Comparing (1) and (2) we then have $|G_f|=s$ and the face is regular.
Otherwise $|g_e|=1$. Again comparing (1) and (2) we have $|G_v|=\frac{\rho}{2} \text { and } |G_f|=\frac{s}{2}$. Therefore $\rho\ge  4$ and $s\ge 4$ and then Euler's formula gives $$\frac{E}{2}+\frac{E}{2}\ge E+2.$$
If $G$ includes reflections
The only 'extra' case is $|G_v|=\rho, |G_e|=2, |G_f|=s$, where both $G_v$ and $G_f$ contain reflections. However, the orders of the two groups, $\rho$ and $s$, are then both even and we still have $\rho\ge  4$ and $s\ge 4$.
A: This is a more elaborate version of my comment below the question.
The first two observations are:

*

*if $P$ is edge-transitive, then all its edges are of the same length.

*if $P$ is vertex-transitive, then $P$ is inscribed, that is, all its vertices are one a common sphere.

The next observation is the following: if $P$ has both properties, then also all its faces must have both properties. This is obvious for the edge lengths. For being inscribed consider the following image:

All vertices of $P$ are on a sphere $S$. If $f$ is a face of $P$ and $E$ is the plane that contains $f$, then $S\cap E$ is a circle that contains all the vertices of $f$. In other words, $f$ is inscribed as well.
So all faces of $P$ are inscribed and have all edges of the same length.
The final observation is that the faces must then be regular polygons.
This is easiest seen via an explicit construction:

If the radius of the circle and the edge lengths are fixed, then placing a single edge in the circle inductively determines all other edges as shown in the figure. That is, the inscribed polygon with this edge length is uniquely determined. But a regular polygon has this property, and so the face must be this regular polygon.
A: I shall use Euler's polyhedron formula. (It does generalise to a higher dimension!)  Let each face have $e$ edges and let $\rho$ edges meet at each vertex and then we have the standard result that $$\frac{1}{\rho}+ \frac{1}{e}>\frac{1}{2} \text { and therefore one of } e,\rho \text { is } 3.$$ If $e=3$ then we have triangular faces with edges of equal length and we are finished.
Therefore $\rho=3$. If the three angles at a vertex are not all equal the ratio of different sized angles is either $1:2$ or $1:1:1$. This ratio must be the same for the angles in a face and so $e$ is a multiple of $3$ and then $\frac{1}{\rho}+ \frac{1}{e}$ is too small.
