= 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 are
$$
\boxed{I,R,R^2,R^3,R^4,\qquad m,Rm,R^2m,R^3m,R^4m.}
$$
Every nonidentity rotation has order five. The reflections are the five order-two elements, since $(R^jm)^2=R^jR^{-j}=I$.
The reflection $R^jm$ 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 $j-k\pmod5$. 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.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2-2014-ib-pentagon-reflections.png]
{title=The five order-two pentagon symmetries; each dashed reflection axis fixes one vertex and exchanges the two indicated pairs}
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