Polytopes with equal facets My question is very simple though I was not able to find any related information.

Is it true that if a convex polytope has combinatorially isomorphic facets then it is combinatorially isomorphic to a regular polytope?

I am especially interested in dimensions $3$ and $4$. I think that it is true for dimension $3$ for some simple reason and not sure about dimension $4$.
Thanks in advance.
 A: No, that's not true. Any bipyramid (in any dimension) over a regular polytope has this property, but generally isn't regular itself. So that gives you infinitely many examples in dimension 3; the $n$-gonal bipyramids have triangles for all their facets, but aren't regular for $n=3$ or $n > 4$.
You also have the 13 Catalan solids, or the infinite series of trapezohedra.
Similarly, any simplicial polytopes have all their facets isomorphic, but aren't combinatorially regular.
This property is known as being combinatorially isotopic, or facet-transitive. (Actually, that's a stronger property: not only are all the facets isomorphic, but the polytope's own automorphisms carry each facet onto any other.)
A: Adding simplicity from your comment onto your task, then you'll have both: alike (d-1)-facets and alike 0-facets (vertices). This usually is called a noble polytope. 
Even with the additional restriction to all unit edges, regular polygonic faces, etc. this small non-exhaustive listing shows that not all 4D noble polytopes are regular.
--- rk
A: The smallest counterexample that comes to my mind is the rhombic dodecahedron.

