Given positive integers $d$ and $v$ with $v \geq d+1$, does there always exist a (convex) vertex-transitive $d$-polytope with $v$ vertices? It seems that the answer should be "obviously" true, but I don't know of any natural constructions. A couple of points:


1) If the question is changed to the existence of regular polytopes (i.e. polytopes that are transitive with respect to all the faces), then the answer is 'no' at least for arbitrary $v$.


2) For $d = 2$ the answer is trivially 'yes' by taking the convex hull of $v$ equi-spaced points on the circle.



Vertex-transitive polytopes in any dimension with any number of vertices?
There is no three dimensional vertex transitive polyhedron with 7 vertices. The rotation groups of three dimensional polyhedra are finite quaternion groups. The only finite quaterionion groups whose order is divisible by 7 are associated with polygons with 7 or a multiple of 7 sides. So there have to be another element in the group to move points out of the plane and this result in more than 7 points. There is a complete classification of finite isometric groups that leave one point fixed in dimensions 3 and 4.


There are also no three dimensional vertex transitive polyhedra with 5 vertices but in that case there are more finite quaternion groups whose order is divisible by 5. When these are examined there are no three dimensional vertex transitive polyhedra but there is one in four dimensions namely the four dimensional simplex. 


