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    Assume that $x_1,x_2,\dots,x_{n-1}$ are fixed points of a simple $n$-fold rotation $\rho$. Prove that $\rho$ is a rotation if and only if $\sum\limits_{i=1}^{n-1} (x_i-x_{i-1})=0$.

    So in the case of a circle, $x_i=r\cos(2\pi i/n)$ and $r$ is a fixed radius. In this case, $\rho^n=1$, so $\rho$ is a rotation. I’m having a little trouble getting started.
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