Past exam of the options I course of the University of Oxford / 2010 / B (Group Theory)

Burnside's lemma gives the number of orbits asFor $m=2n-1\ge 3$, the polygon symmetry group contains the $m$ rotations through angles $2\pi j/m$, $0\le j<m$, and $m$ reflections. Each reflection axis passes through a vertex and the midpoint of the opposite edge. These are all the symmetries, so $|G|=2m=\boxed{4n-2}$.

For the pentagon, apply the same group element simultaneously to vertex colours and edge colours. The identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations is a 5-cycle on vertices and also on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two swapped vertex pairs, and one fixed edge and two swapped edge pairs. It therefore fixes $v^3{e}^3$ colourings. Averaging yieldsThe two colour palettes are chosen independently, but rotations and reflections act on both parts of the same pentagon.