= Solution
<Burnside's lemma> gives the number of orbits as
$$
|\Omega/G|=\frac1{|G|}\sum_{g\in G}|\operatorname{Fix}(g)|.
$$
For $m=2n-1\geq3$, the polygon symmetry <group> contains the $m$ rotations through angles $2\pi j/m$, $0\leq j<m$, 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 $v^5e^5$ 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 $v^3e^3$ colourings. Averaging yields
$$
\boxed{\frac{v^5e^5+4ve+5v^3e^3}{10}.}
$$
The two colour palettes are chosen independently, but rotations and reflections act on both parts of the same pentagon.
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