For this question, a "cycle" is a sequence of distinct points $X = (x_1,x_2,\cdots,x_k)\in\mathbb{R}^3$ which defines a piecewise linear path starting at $x_1$ and visiting the points in order and returning to $x_k$ (below we use the cyclic indexing notation $x_{k+1}=x_1$ and $x_0=x_k$).
Given a cycle $X$, the "velocity variables" are the unit vectors $$ v_j = \frac{x_{j+1}-x_j}{|x_{j+1}-x_j|}. $$ At each point in the cycle we then define the normal vector $$ n_j = \frac{v_j - v_{j-1}}{|v_j-v_{j-1}|}. $$$$ n_j = \frac{v_j - v_{j-1}}{|v_j-v_{j-1}|} $$ and the angle of reflection $$ \theta_j = \cos^{-1}(-v_j\cdot v_{j-1}) $$
The interpretation is as follows: the pair $(x_j,n_j)$ represents a small mirror placed at the point $x_j$ and oriented to have normal vector $n_j$. Then the cycle $(x_1,x_2,\cdots,x_k)$ is a valid trajectory of a beam of light exiting at $x_1$ along the direction $v_1$ and bouncing off each mirror in order and then returning to $x_1$. The angles $\theta_j$ are the angles between the outgoing and incoming rays.
The question is as follows: can there exist a nontrivial permutation of the cycle which keeps the normal vectors $n_j$ and the angles $\theta_j$ fixed? In other words, is it possible to set up a finite set of mirrors that admits two distinct periodic trajectories that each hit all of the mirrors with the same angles of reflection?