Action of the symmetry group on polygons

Question

I want to show that a polygon $P$ is regular, i.e. all the angles are equal, iff $\Sigma(P)$, the symmetry group, acts transitively on Vert$(P)$.

Answer 1

The term equiangular polygon (Wikipedia) is preferred to describe the condition that all (interior) angles of a polygon are equal. To be regular a polygon is conventionally defined to be both equiangular and equilateral (all sides of equal length).

It is easy enough to show that if the symmetry group of a polygon is vertex transitive then all interior angles are equal. Indeed if some symmetry of the polygon maps vertex $V$ to vertex $V'$, then that map establishes the equality of the angles at $V,V'$.

The other direction is not true. An equiangular polygon need not have a vertex transitive symmetry group. Indeed something of the sort was already shown by the earlier Question Are all equiangular odd polygons also equilateral? The example described there of an equiangular pentagon which is not equilateral perturbs a regular pentagon so that instead of having all sides equal, one gets an equiangular pentagon with bilateral symmetry but not a vertex transitive symmetry group.

Indeed one might as easily have an equiangular pentagon with all five edge lengths different, so that the symmetry group is trivial and thus vertex transitivity is lacking. Wikimedia contributor J Hokkanen posted an image of such a pentagon (licensed CC A-S-A 3.0):

It is possible to introduce additional assumptions in order to draw the conclusion of vertex transitive symmetry (or even regularity) from equiangularity. The Wikipedia article linked above might suggest some ideas of this kind, and if you are interested I can elaborate.