Because $A_iA_{i+1}$ is parallel to $B_iB_{i+1},$ by arguments of lines, we must have $\angle A_iA_{i+1}A_{i+2}=\angle B_iB_{i+1}B_{i+2}.$ Now, fix the circumcircle of $93$-gon $B.$ We prove that $B$ must also be fixed. This is because, $\angle B_iB_{i+1}B_{i+2}$ is fixed, thus the arclength of $B_iB_{i+2}$ is fixed. Now, the arclengths of $B_1B_3, B_3B_5, \dots$ must be fixed. Thus, the arclength of $B_{93}B_1$ is fixed. Similarly, any other arclength must also be fixed. Thus, the $93$-gon $B$ must be fixed, if its circumcircle is fixed. However, the polygon similar to $A$ inside the circumcircle satisfies the conditions, and thus must be $B.$ This means $A$ and $B$ are similar and we are done.

Motivation: Our goal is essentially to prove they are similar. So, we want a contradiction if we tweak the positions of, say $B_1$ slightly. If the arclength of $B_1B_2$ increases, that of $B_2B_3$ decreases, $B_3B_4$ increases, and so on. To use this, we may note the arclength of $B_1B_3$ is fixed. Another way is to note that $B_{93}B_1$ must increase, so $B_1B_2$ must decrease, which is a contradiction.