For $n=3$ the answer is clearly $0$. For $n=4$ a quadrilateral with four equal sides is a rhombus which is convex, so the answer is $0$ again. From now on let $n \ge 5$. The sum of interior angles is $(n-2) \cdot 180^\circ$, so there can be at most $n-3$ of these concave angles. We shall see that this is indeed possible: For odd $n=2k+1$, consider a triangle with sides $k,k,1$. We slightly deform the $k$ segments so that they become a bit longer and form an arc inside the original triangle. Then we also make the short side of the triangle slightly larger. If we do this properly, we can achieve that all sides still have the same length, we still don't have any self-intersections, but we now have $2(k-1)=n-3$ concave angles. For even $n=2k+2$ with $k \ge 2$ we do the same with a triangle of sides $k,k,2$ (which exists since $k \ne 1$). So the answer is $0$ for $n=4$ and $n-3$ for $n \ne 4$.