Dihedral group - elements of order $2$

If $D_{n}$ is the Dihedral group of degree $n$ (order $2n$) where n is an odd number, can I then conclude that the numbers of elements of order 2 in $D_{n}$ is equal $n?$

I suppose there are the following types of elements in $D_{n}$: b, $a^{i}$, $a^{i}b$ for some $0\leq i < n$ .

$\bf{Convention:}$ 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.

recall that $ba = a^{-1}b$. let's prove that $ba^k = a^{-k}b$ for all $k$, by induction on $k$:

we have the base case given above. by an induction hypothesis, we have:

$ba^{k-1} = a^{-k+1}b$, and so:

$ba^k = (ba^{k-1})a = (a^{-k+1}b)a = a^{-k+1}(ba) = a^{-k+1}(a^{-1}b) = a^{-k}b$, QED

we can use this to show that the order of $a^kb$ is $2$:

$(a^kb)^2 = (a^kb)(a^kb) = a^k(ba^k)b = a^k(a^{-k}b)b = b^2 = e$

we have exactly $n$ choices for $k$, and no element of $\langle a \rangle$ is of order $2$ (since $n$ is odd), so we have exactly $n$ elements of order $2$.