Say we have $n$ children sitting in a ring. Each child is given an arbitrary toy that is either red or blue. We know that children, being as they are, can be somewhat fickle and very sociable, so if a child does not like their toy, they swap toys with their immediate neighbor. Neighbors will only trade toys once. They can swap toys of the same color (a red firetruck for a red yo-yo, for example), or toys of different colors. After the trades are done, we then go back around the circle and check the colors of each toy. We finally say that two colorings of the graph $C_n$ are isomorphic if one can be achieved after exactly $k$ trades. For example, in the diagram below, the three colorings in the green box are all isomorphic to the leftmost graph with $k=1$, while the one outside the box is not.
The question then arises: for $n$ children and $k$ swaps, how many unique distributions of toy colors are there? My first instinct was to try using PET to solve this, but I quickly ran into an issue; the theorem only works with groups, and the set of permutations here is not a group. For example, again, with $k=1$ and $n=4$, the permutations $(12)$ and $(34)$ are both members of the set, but $(12)(34)$ is not. It is not a closed set.
My next instinct was to use PET on each individual permutation and then add them up. For example, I thought we could say $G=\{I,(12)\}$, and then $Z(G)=\frac{1}{2}(a_4+a_1^2a_2)$, and then rinse and repeat for our other permutations. However, again, we run into a problem. With this method, multiple versions of the same graph get counted as "unique" graphs. This method is a no-go as well.
How can I approach this problem? I feel like PET or Burnside's lemma should be applicable here to sort out all the duplicated, but I can't for the life of me figure out how.