= Solution
Use the <dihedral group> presentation $G=\langle r,s:r^{2n}=s^2=1,\ srs=r^{-1}\rangle$. Every element is $r^j$ or $sr^j$. A nonidentity rotation has order two only when its exponent is $n$, while $(sr^j)^2=1$ for every $j$. Thus the complete list is
$$
\boxed{r^n,\quad sr^j\ (0\le j<2n).}
$$
The half-turn $r^n$ is central, so its <conjugacy class> is the singleton $\{r^n\}$ and its <normal closure> is \b[$\{1,r^n\}$].
For reflections, $r^k(sr^j)r^{-k}=sr^{j-2k}$ and $s(sr^j)s^{-1}=sr^{-j}$. Hence the <conjugacy class> is exactly the $n$ reflections whose exponents have the same parity as $j$:
$$
\boxed{\{sr^{j+2k}:0\le k<n\}.}
$$
The <subgroup> generated by this class contains $r^2$, as the product of two consecutive listed reflections, and contains $sr^j$. Conversely those two generators contain every listed reflection. Their <subgroup> is invariant under <conjugation> by both $r$ and $s$, so the smallest <normal subgroup> containing $sr^j$ is
$$
\boxed{\langle r^2,sr^j\rangle
=\{r^{2k},sr^{j+2k}:0\le k<n\},}
$$
of order $2n$ and index two. This <involutions in an even dihedral group> classification distinguishes the two reflection classes from the central half-turn.
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