 Solution

For a regular pentagon, take $R$ as rotation through $2\pi /5$ and $m$ as reflection in the horizontal axis through vertex zero. The distinct elements areEvery nonidentity rotation has order five. The reflections are the five order-two elements, since $(R^{j}m{)}^2=R^j{R}^{-j}=I$.

The reflection $R^{j}m$ has axis at angle $j\pi /5$ to the horizontal, understood modulo $\pi$. If vertices have angles $2k\pi /5$, it sends vertex $k$ to vertex $j-k(\mathrm{mod}5)$. Each axis passes through one vertex and the midpoint of the opposite edge; that vertex stays fixed and the other four vertices exchange in two pairs.