Assume that there are $k\leq n$ points on the boundary of the convex hull of the set. For each of the inner points we have that the minimum angle with vertex on that point is $\leq \frac{2\pi}{n-1}$ (since there are $n-1$ angles with sum $2\pi$. If a point on the boundary has inner angle $\alpha$ the minimum angle with vertex on that point must be $\leq\frac{\alpha}{n-2}$ because there are $n-2$ consecutive angles with that vertex. Finally the least of the interior angles of the convex hull must be $\leq \frac{(k-2)\pi}{k}$. Therefore the minimum angle must be $\leq min\{\frac{2\pi}{n-1}, \frac{(k-2)\pi}{k(n-2)}\}=\frac{(k-2)\pi}{k(n-2)}\leq\frac{(n-2)\pi}{n(n-2)}=\frac{\pi}{n}$. The equality can only be achieved if all the points are on the boundary of the convex hull, and also the interior angles must be all equal, and the angles with a same vertex must also all be equal. Fixing two consecutive vertices $A_1,A_2$ all angles $\angle A_1A_iA_2$ must also be equal to $\frac{\pi}{n}$ and so the polygon is cyclic and $\angle A_1OA_2=\frac{2\pi}{n}$ where $O$ is the circumcenter of the polygon, so all chords must be the same length and thus the set must form a regular $n$-gon.

Just wanted to add one small detail about the inequality at the end (though I might be just not seeing something). We have \begin{align*} \frac{k-2}{k(n-2)} &\leq \frac{k}{(n-2)k} - \frac{2}{(n-2)k} \\ &\leq \frac{1}{n-2} - \frac{2}{(n-2)n} \tag{since $k \leq n$} \\ &= \frac{1}{n} \end{align*}