Since the set of paths is isomorphic to \$$\mathbb{N}\$$, let's see what the equation means in \$$\mathbb{N}\$$. \$$s^4=s^2\$$ corresponds to \$$n\equiv n-2\$$ for \$$n\geq4\$$. This means that every even number except 0 is equivalent to 2, and every odd number except 1 is equivalent to 3. So, there are 4 morphisms in \$$\mathcal{D}\$$: \$$\mathrm{id}_z,s,s^2,\$$ and \$$s^3\$$.

Fredrick Eisele wrote in [#1](https://forum.azimuthproject.org/discussion/comment/18569/#Comment_18569):
>I believe the morphisms form an equivalence group over the paths.

They certainly form equivalence classes. They don't quite form a group, though the paths of length at least 2 have a \$$\mathbb{Z}/2\mathbb{Z}\$$-like structure (with composition corresponding to addition modulo 2, and the even numbers corresponding to the identity).