$\bf{Convention:}$ Let $n$ be an odd number greater that or equal to $3$. Let $D_n$ denote the group of symmetries of regular $n-$ gon. Note that $|D_n|=2n.$
Yes, you're right. The elements of order $2$ in the group $D_{n}$ are precisely those $n$ reflections. Note that these elements are of the form $r^ks$ where $r$ is a rotation and $s$ is the reflection; $1 \leq k \leq n$.
In fact, we get some information about the Sylow Structure of $D_n$. Since, the number of elements of order $2$ in a group is the number of subgroups of order $2$, and that its exponent in $|D_n|$ is unity, we have that, the number of Sylow $2-$ subgroup of $D_n$ is $n$.
Now, the inquisitive$^\dagger$ reader would have observed that: For each $n$ such that $n \equiv 1 \bmod 2$, there is a group with exactly $n$ Sylow $2-$ subgroups.
Now, one can more generally ask:
Given a prime $p$, is there a group such that there are exactly $n$ Sylow $p-$ subgroups whenever $n \equiv 1 \bmod p$?
The even case of this problem is not hard to answer either: Note that the reflections are always of order $2$. So, for a regular $n-$ gon, with $n$ even, we have another element of order $2$-Rotation of the polygon by $\pi$. Note that these are the only elements of order $2$. This is done algebraically, via presentations, in Dylan's answer.
$\dagger$ It should go without saying (?) that I was not inquisitive and I realised I could ask this question only after reading one of Keith Conrad's blurbs. I cannot fish that link out now, but will add whenever I found that out. I am also reasonably sure, there are references where some examples for which the answer is negative has been discussed. My hearty thanks to him for leaving a commment here.