Nice answers, Reuben!

> Questions: is the notion of "interpretation" I'm informally using formalizable as a functor?

Yes! If you have a category \\(\mathcal{C}\\) with one object and with all morphisms invertible, it's secretly just a group. And if you have a functor \\(F: \mathcal{C} \to \mathbf{Set}\\), it gives a set and an action of your group as permutations of that set.

For example, your set could be the 4 corners of the square, or the set of all points in the square. Your category \\(\mathcal{C}\\) could be the one in Puzzle 107. The functor \\(F : \mathcal{C} \to \mathbf{Set}\\) sends the one object \\(z\\) in \\(\mathcal{C}\\) to your set. It sends the four morphism \\(1_z, s, s^2, s^3\\) in this category to the four rotations of your set!

More generally, for any \\(\mathcal{C}\\) , a functor \\(F: \mathcal{C} \to \mathbf{Set}\\) gives you a bunch of sets (one for each object in your category) and functions (one for each morphism) obeying the same relations as the morphisms in your category. So, it gives a "concrete picture" of the abstract category \\(\mathcal{C}\\).

I think this all I will say now, but there's a lot more to say here.