This is the graph I am working with and the edge labelings are given by $l(\{2n, 2n + 1\}) = a$ and $l(\{2n + 1, 2n + 2\}) = b$.
The question I'm working on guides us through to a description of the automorphism group of this graph. I so far have the isomorphisms (defined on the vertices) given by $\mu (n) = 1-n$ and $\nu (n) = n+2$. And I have that $\nu^k (n) = n + 2k$ for all $k \in \mathbb{Z}$. I have also shown that $\nu^{-m}(\{2m, 2m+1\}) = \{0,1\}$ for all $m \in \mathbb{Z}$. We are told that $\phi$ is an element of Aut($\Gamma$).
Now, the part I am confused about is as follows.
Define $\theta = \nu^{-m} \circ \phi$, where $\phi$ is as above, so $\theta (\{0,1\}) = \{0,1\}$. Show that either $\theta = \text{Id}_{\Gamma}$ or $\mu \circ \theta = \text{Id}_{\Gamma}$. Use this to show that either $\phi = \nu^m$ or $\phi = \nu^m \circ \mu$. Thus, show that $\text{Aut}(\Gamma) = \{\nu^m, \nu^m\mu : m \in \mathbb{Z}\}$ and that no two elements of the latter set are equal.
I honestly don't know how to tackle this at all, everything I try seems to just go in circles and I don't think putting my workings here would help. I feel like I don't have enough information about $\phi$, but I must do... I think maybe I'm just struggling to comprehend what the question is asking?
I have basic graphs knowledge (this is the first and only question I've done involving edge labeling and automorphisms), and fundamental group theory knowledge.
This is what I've got, and I think it makes sense...
There are two cases for $\theta$. \begin{equation} \text{Case 1: }\theta (0) = 0 \text{, and } \theta(1) = 1 \\ \text{Case 2: }\theta(0)=1 \text{, and } \theta(1) = 0 \end{equation}
Given case 1, I claim that $\theta = \text{Id}_{\Gamma}$.
As $\theta$ is an isomorphism, case 1 forces $n \mapsto n$, $\forall n \in \mathbb{Z}$. To illustrate, consider where $\theta$ sends $2$. It cannot be that $2 \mapsto 0$ nor $2 \mapsto 1$ as then $\theta$ would not be injective. Additionally, as $\theta$ is an isomorphism, the vertex $2$ must be incident to the vertex $1$ in the image of $\theta$. Thus, we must have $2 \mapsto 2$, and similarly for the vertex $-1$ and so on. Therefore, $\theta = \text{Id}_{\Gamma}$.
Given case 2, I claim that $\mu \circ \theta = \text{Id}_{\Gamma}$.
Notice that, $\mu(\theta(1)) = 1$ and $\mu(\theta(0)) = 0$. And so, the argument above applies in this case to $\mu \circ \theta$, i.e., $\mu \circ \theta = \text{Id}_{\Gamma}$.