Number of $p$-Sylow subgroups in $D_{2n}$

Question

Let $2n = 2^ak$, where $k$ is odd. We wish to show that the number of $2$-Sylow subgroups of $D_{2n}$ is $k$. My approach has been to construct such a $2$-Sylow subgroup $P_2$, and then show that the size of the stabilizer of one of its elements (under the conjugation group action) is $2^a$, which I believe would prove the result since $|D_{2n}|/|\text{stab}_{D_{2n}}(x)| = \frac{2^ak}{2^a} = k =|\text{orb}(x)| $, and the orbit of $x$ is in bijection with the set of all $2$-Sylow subgroups since they are all conjugate. I am, however, running into some trouble doing this, and after computing some explicit examples it seems like my approach is incorrect. Where did I go wrong? Is the orbit of $x \in P_2$ not in bijection with the set of all $2$-Sylow subgroups?