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).

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).