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.