Here's a very interesting problem that I made up with a friend this morning:
For which even $n$ does there exist a permutation $\pi$ of $\{1,2,\cdots,n\}$ such that when we reflect any point $P$ in the $xy$ plane about the lines $y=\pi(1)x, y=\pi(2)x, \cdots, y=\pi(n)x$ in that order, $P$ is returned to it's original position? For these $n$, how many such permutations are there?
So far, we've noted the fact that when we reflect a point about two lines, it's equivalent to a rotation about the intersection of the lines of angle twice the angle between the two lines. Thus, if we let $\{a_1,a_2,\cdots,a_{n/2}\}=\{\pi(1),\pi(3),\cdots,\pi(n-1)\}$ and $\{b_1,b_2,\cdots,b_{n/2}\}=\{\pi(2),\pi(4),\cdots,\pi(n)\}$, the image of $P$ is a rotation of angle $2\left(\sum \arctan(a_k)-\sum \arctan(b_k)\right)$. This needs to be divisible by $2\pi$. With $\arctan(t)=-\arctan(-t)$ and some basic facts about complex numbers, this is equivalent to requiring $\prod (1+a_ki)\prod (1-b_ki)\in\mathbb{R}$. From here, there are two main paths that I tried. The first is using that $z\in\mathbb{R}\iff z=\overline{z}$. This doesn't get far. The second is trying to find the imaginary part and setting equal to $0$. I couldn't finish this approach off either. I wrote a C++ program and found that this is impossible for $n=4,6,8,10,12,14$. Any ideas?