Place the diagram on the complex plane. Let $w$ be the unit circle. Let $c=e^{\pi i/n}$, and $A_i$ be $w^i$. Let $P$ correspond to $p$. Let $H_{n+1}=H_1$. We are given that $H_i$ is the orthocenter of $\triangle PA_{2i-1}A_{2i}$. Since this triangle will always be inscribed in the unit circle, we find that $H_i$ corresponds to $P+A_{2i-1}+A_{2i}=p+c^{2i-1}+c^{2i}$. We know $$H_iH_{i+1}=|H_{i+1}-H_i|=|p+c^{2i+2}+c^{2i+1}-(p+c^{2i-1}+c^{2i})|=|c^{2i+2}+c^{2i+1}-c^{2i-1}-c^{2i}|.$$This is equal to constant $|c^{2i}||c^2+c-1-1/c|=|c^2+c-1-1/c|$, so $H_iH_{i+1}$ is constant for all $i$. Thus, $H_1H_2 \dots H_n$ is a regular $n$-gon.

Since the distance between the vertex and orthocenter is twice the distance between circumcenter and opposite side so $PH_i = 2OM_i$ where $M_i$ is the midpoint of $A_{2i-1}$ and $A_{2i}$ and O is circumcenter of $A_1 \dots A_{2n}$ $\therefore$ $H_1 \dots H_n$ is concyclic with $P$ as the center of the circumcircle. Now we find the $\angle H_i P H_{i+1}$ We drop perpendiculars from P on the sides $A_{2i-1} A_{2i}$ and $A_{2i+1} A_{2i+2}$ Then let $\angle A_i O A_{i+1} = \alpha$ Then, $\angle A_{2i-1} P A_{2i+1} = \alpha$ Let $P A_{2i} A_{2i-1} = \beta$, then $P A_{2i+2} A{2i-1} = \beta$ Again, $H_i P A_{2i-1} = 90 - \frac{\alpha}{2} - \beta$ $A_{2i+1} P H_{i+1} = \frac{\alpha}{2} -(90 - \alpha - \beta)$ So adding all of these we get $\angle H_i P H_{i+1} = 2\alpha$ So the polygon is regular

eduD_looC wrote: Place the diagram on the complex plane. Let $w$ be the unit circle. Let $c=e^{\pi i/n}$, and $A_i$ be $w^i$. Let $P$ correspond to $p$. Let $H_{n+1}=H_1$. We are given that $H_i$ is the orthocenter of $\triangle PA_{2i-1}A_{2i}$. Since this triangle will always be inscribed in the unit circle, we find that $H_i$ corresponds to $P+A_{2i-1}+A_{2i}=p+c^{2i-1}+c^{2i}$. We know $$H_iH_{i+1}=|H_{i+1}-H_i|=|p+c^{2i+2}+c^{2i+1}-(p+c^{2i-1}+c^{2i})|=|c^{2i+2}+c^{2i+1}-c^{2i-1}-c^{2i}|.$$This is equal to constant $|c^{2i}||c^2+c-1-1/c|=|c^2+c-1-1/c|$, so $H_iH_{i+1}$ is constant for all $i$. Thus, $H_1H_2 \dots H_n$ is a regular $n$-gon. I think you haven't completed your solution. Since rhombus is not a regular polygon, so by proving sides are equal doesn't necessarily prove that it is a regular polygon. Sorry for any confusion. Thank you.