I was working with $\mathbb{Z}/6\mathbb{Z}$ acting on a regular hexagon where $\bar{n}$ acts as rotation by $2n\pi/3,$ clockwise. By labeling the vertices (I started with $1$ at the top left corner going clockwise) of the hexagon $1,2,3,4,5,6$ we have the homomorphism $\alpha: \mathbb{Z}/6\mathbb{Z} \rightarrow S_6$ such that $$\bar{0},\bar{3}\mapsto (1)$$
$$\bar{1},\bar{4}\mapsto (1\:2\:3\:4\:5\:6)$$
$$\bar{2},\bar{5}\mapsto (1\:3\:5)(\:2\:4\:6)$$
Now, let $O$ be the set of opposite vertices i.e. $O=\{\{1,4\},\{2,5\},\{3,6\}\}.$ Then we have the homomorphism $\beta: \mathbb{Z}/6\mathbb{Z} \rightarrow S_O$ such that
$$\bar{0},\bar{3}\mapsto\{\{1,4\},\{2,5\},\{3,6\}\}$$
$$\bar{1},\bar{4}\mapsto\{\{2,5\},\{3,6\},\{1,4\}\}$$
$$\bar{2},\bar{5}\mapsto\{\{3,6\},\{1,4\},\{2,5\}\}$$
Is this correct?
To clear any ambiguity the following is how I labeled my hexagon:
Best Answer
You have the right idea, but your calculations are off.
With your current labeling, the map $\mathbb{Z}/6\mathbb{Z}\to S_6$ you've defined has the following images. I write '$e$' to be the identity permutation.
$$\begin{array}{ll} \bar{0} & \mapsto \text{ rotation by 0 } & = e \\ \bar{1} & \mapsto \text{ rotation by }\frac{2\pi}{3} & = (135)(246) \\ \bar{2} & \mapsto \text{ rotation by }\frac{4\pi}{3} & = (153)(264) \\ \bar{3} & \mapsto \text{ rotation by }2\pi & = e \\ \bar{4} & \mapsto \text{ rotation by }\frac{8\pi}{3}=2\pi + \frac{2\pi}{3} & = (135)(246) \\ \bar{5} & \mapsto \text{ rotation by }\frac{10\pi}{3} = 2\pi + \frac{4\pi}{3} & = (153)(264) \\ \end{array}$$
With this, can you find the images of each $\bar{n}$ for the map $\mathbb{Z}/6\mathbb{Z}\to S_O$?