A regular polygon is a bounded, not self-intersecting polygon in which all edges have the same length and all interior angles are identically, say, $\theta$. It is known that $\theta=(1-2/n)\pi$ for some $n\ge 3$.
Now, define a pseudo-regular polygon as a bounded, not self-intersecting (but not necessarily convex) polygon in which all edges have the same length and all interior angles are $\in\{\theta,2\pi-\theta\}$ for some fixed $\theta<\pi$. All regular polygons are pseudo-regular, and there are others:
Question: Are there any with $\theta\not=(1-2/n)\pi$ for all $n\ge 3$?
Update
Based on the comment of nickgard, I found a solution and posted it as an asnwer. It was surprisingly straight forward.
No, there aren't any others.
The interior angle sum of a bounded, not self-intersecting but not necessarily convex $n$-gon is always $(n-2)\pi$. Now, if there are $n_1$ vertices with interior angle $\theta$, and $n_2$ vertices with interior angle $2\pi-\theta$, we obtain
$$n_1\theta+n_2(2\pi-\theta)=(n_1+n_2-2)\pi.$$
This rearranges to
$$\theta=\Big(1-\frac2{n_1-n_2}\Big)\pi,$$
and so we see that $\theta$ must be the interior angle of a regular $n$-gon.
Update
I just realized that this this answer assumes that the pseudo-regular polygon is connected and simply connected (i.e. the boundary is connected). Here is an example with $\theta=\pi/2$ that is not simply connected.
Anyway, my argument still holds. One just has to apply the argument to each connected component and to each hole separately.