Permutations Given by Rotating Points in the Plane

Question

Consider a set $X=\{p_1,p_2,p_3,...,p_n\}$ points in $\mathbb{R}^2$. Now consider the sets $R(X)$ for $R$ rotation matrices. For all but finitely many $R$, projecting $R(X)$ onto the $x$-axis gives an bijection between $R(X)$ and its image, and therefore corresponds to a permutation of $X$. For simplicity, let us assume that if $R$ is the identity, then the points $X$ are in order, that is $p_1$ projects onto a lower value that $p_2$ which projects onto a lower value that $p_3$, and so on. The the sets of orderings of $X$ induced by rotations corresponds to a subset $Y$ of the symmetric group $S(X)$ on $X$. My question is, what are the possible sets $Y$? $Y$ must contain $\{p_n,p_{n-1},p_{n-2},...,p_1\}$ simply because of rotation by $\pi$ radians. Similarly, if $n>1$ is even then $Y$ is even. Is it possible to have $Y=S(X)$?

Answer 1

Not a complete characterization, but in general we can say the following. At each of the finite set of rotations at which the order changes, $2$ or more points change order. All of the points whose order change at a specific such rotation must be collinear, they change order when the line passing through them is parallel to the $y$-axis, and they reverse order, then reverse order again after a further $180^{\circ}$ rotation. This means the number of such rotations is at most $2 {n \choose 2}$ (twice the number of pairs of points, since a pair of points determines a line), so at most $2 {n \choose 2}$ different orders can occur.

This means it's not possible to have $Y = S(X)$ for $n \ge 4$ since $2 {n \choose 2} < n!$ for all such $n$. However it is possible for $n = 2$ (of course) and for $n = 3$ (a triangle whose interior doesn't contain the origin should work).