Here is the question:
Suppose $\varphi\colon\Bbb R^2\to\Bbb R^2$ is an isometry and $\varphi(\pi_n)=\pi_n$, where $\pi_n$ is the regular $n$-gon with center at origin. Why must $\varphi$ fix the origin or the center?
Here, I don't want to use the fact that $\varphi$ is a permutation of vertices of $\pi_n$, because it is actually equivalent to the question I asked, and I don't know its proof either.
I think this question is really easy, because intuitively it is correct. However, I want to prove it rigorously and don't know how to proceed.
Partial Answer:
I came up with a proof for $n=2q$ and am not sure if it is correct.
If $\varphi$ is an isometry and $S[A,B]$ is a line segment between $A\in\Bbb R^2$ and $B\in\Bbb R^2$, then $\varphi(S[A,B])=S[\varphi(A),\varphi(B)]$ and the length is also preserved. In particular, the midpoint of $S[A,B]$ is mapped to the midpoint of $S[\varphi(A),\varphi(B)]$. Take two vertices $v_i,v_j$ such that $O\in S[v_i,v_j]$ and consider four facts: (1) Every regular polygon has a circumscribed circle. (2) $S[v_i,v_j]$ is the diameter of the circumscribe circle. (3) Vertices of $\pi_n$ are on the circle. (4) $\varphi(\pi_n)=\pi_n$
Combining the four facts above, $\varphi(v_i)$ and $\varphi(v_j)$ must also be vertices of $\pi_n$, so that the midpoint of $S[\varphi(v_i),\varphi(v_j]$ is the center $O$. Therefore, the origin $O$ is preserved.
Approach 1: Circumcenter.
Approach 2: Incenter.