Let $n$ be a positive integer number and $P$ be a point in a plane. Let $A\_1$, $A\_2$, $\cdots$, $A\_m$ be $m$ points in the plane, we take modulo $m$ for $A\_j$ (it is mean $A\_{m+i}=A\_{i}$ for $i=1, 2, \cdots$).


Now we [rotate](https://en.wikipedia.org/wiki/Rotation_(mathematics)) $P$ clockwise above $A\_1$ with angle $\frac{\pi}{n}$ we have point $P\_1$, rotate $P\_1$ clockwise above $A\_2$ with angle $\frac{\pi}{n}$ we have point $P\_2$ $,\cdots,$ rotate $P\_k$ clockwise above $A\_{k+1}$ with angle $\frac{\pi}{n}$ we have point $P\_{k+1}$....


By my computation with $(m,n)=(3,1), (3,2), (3,3), (4,3), (3,4), (2,3), (2,4), (2,5)$....I see that exist $k$ such that $P\_{k+1}\equiv P$


**My question:** What condition of $(m,n)$ such that exist $k$ so that $P\_{k+1}\equiv P$? Version of this problem for Euclidean space (or non-Euclidean geometries)?



Chain rotation of a point
А composition of rotations is ether rotation or translation (a special case of Chasles' theorem) depending on the total rotation angle which is $k\pi /n$, so $P\_k=R\_0^{k\pi /n}(O\_k)$ (rotation with some center $O\_k$) for $k\not\equiv 0\pmod {2n}$ and $P\_k=T\_{v\_k}(P)$ (translation) for $k\equiv 0\pmod{ 2n}$ (This rotation angle can be controled by complex numbers as in Philipp Lampe's comment). So fixed point exists only if $k\not\equiv 0\pmod {2n}$ (this point is $O\_k$) or $k\equiv 0\pmod{ 2n}$ and $v\_k=0$ (all points are fixed). The last case can be checked starting from arbitrary $P$ and checking whether $P\_k=P.$


