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2 added 226 characters in body

I guess you are looking for a nice answer, but here is a stupid one. On the other hand I'm sure that there is no "nice answer".

Say assume your $n$-gons are convex hulls

A polygon can is uniquely determined by length of ${ z_1,z_2,\dots,z_n }sides$\ell_i$and${ w_1,w_2,\dots,w_n}$. You may think parametrize plane by angles$\mathbb C$. You can assume that \alpha_i$. Thus we have to find a complete set of invariants for sequence $\sum w_i=\sum z_i=0$(\alpha_1,\ell_1,\alpha_2,\ell_2,\dots\alpha_n,\ell_n)$which survive after even cyclic shifts and reversing order. Then if polygons are equal then$|\sum z_i^k|=|\sum w_i^k|$you prepare symmetric polynomials for any$k\ge 1$your group. Moreover Say take all monomials of degree at most one in each$(\sum z_i^k)^\ell(\sum w_i^\ell)^k=(\sum w_i^k)^\ell(\sum z_i^\ell)^k$for any \alpha_i$ and $k,\ell\ge 1$\ell_i$and take its mean value it along the group. You can also think that obtain a big collection of polynomial expressions in$k,\ell\le n$\alpha_i$ and $\ell_i$ which gives complete invariant (perimeter will be one of them).

I bet that is enough ;)

1

I guess you are looking for a nice answer, but here is a stupid one. On the other hand I'm sure that there is no "nice answer".

Say assume your $n$-gons are convex hulls of ${ z_1,z_2,\dots,z_n }$ and ${ w_1,w_2,\dots,w_n}$. You may think parametrize plane by $\mathbb C$. You can assume that $\sum w_i=\sum z_i=0$. Then if polygons are equal then $|\sum z_i^k|=|\sum w_i^k|$ for any $k\ge 1$. Moreover $(\sum z_i^k)^\ell(\sum w_i^\ell)^k=(\sum w_i^k)^\ell(\sum z_i^\ell)^k$ for any $k,\ell\ge 1$. You can also think that $k,\ell\le n$.

I bet that is enough ;)