Automorphisms of Cayley graphs of $\mathbb{Z}$

Question

My goal is to show that there are no finite generating sets $A$ and $B$ such that $\mathrm{Cay}(\mathbb{Z},A)$ is isomorphic to $\mathrm{Cay}(\mathbb{Z}\times \mathbb{Z}_2, B)$. My idea for this is to start by noting that $\mathrm{Cay}(\mathbb{Z}\times \mathbb{Z}_2, B)$ has a non-trivial automorphism of order $2$ with exactly two fixed vertices, induced by the map $(a,b) \mapsto (-a,b)$.

Then I have the feeling that there should be no such automorphism of $\mathrm{Cay}(\mathbb{Z},A)$. This is obvious if we pick the generating set $A = \{1\}$; in this case $\mathrm{Cay}(\mathbb{Z},A)$ is just the real line and if a graph automorphism has two fixed vertices then it is trivial. However, this is less clear when the generating set for $\mathbb{Z}$ is $\{n_1, \dots, n_k\}$ with $\gcd(n_1, \dots, n_k) = 1$.

Can anyone provide a proof that an automorphism of $\mathrm{Cay}(\mathbb{Z},A)$ with two fixed points must be trivial, or provide a counterexample? Different approaches to my original problem are also welcome.

Edit. Some clarifications: my question is exactly as formulated in kabenyuk's comment. I want to show that there are no generating sets $A$ of $\mathbb{Z}$ and $B$ of $\mathbb{Z} \times \mathbb{Z}_2$ such that the Cayley graphs $\mathrm{Cay}(\mathbb{Z},A)$ and $\mathrm{Cay}(\mathbb{Z}\times \mathbb{Z}_2,B)$ are isomorphic.

To fix things, here is the definition of the Cayley graph I am working with: for a group $G$ with generating set $S$ not containing the identity, the Cayley graph $\mathrm{Cay}(G,S)$ is the graph with vertex set $G$ and edge set $\{\{ g, gs \} : g \in G , s \in S \cup S^{-1} \}$.

Answer 1

It was proved in [1] that a connected undirected Cayley graph on $\mathbb{Z}$ with finite valency has automorphism group the infinite dihedral group in its natural action. Your claim follows easily.

(By using your approach for example, fixing two points fixes everything.)

[1] Möller, Rögnvaldur G.; Seifter, Norbert. Digraphical regular representations of infinite finitely generated groups. European J. Combin. 19 (1998), no. 5, 597–602. MR1637768 (99i:20007), Zbl 0905.05036, doi:10.1006/eujc.1998.0210.

Answer 2

Suppose $S$ is a generating set of $\Bbb Z$, $G$ is the undirected graph of the Cayley graph of $(\Bbb Z, S)$, and $H$ is the corresponding graph for $(\Bbb Z \times {\Bbb Z}_2, \{(1, 0), (0, 1)\})$. And suppose $G$ and $H$ are isomorphic.

Each vertex of $H$ has degree 4, so $|S| = 2$. Say $S = \{a_0, a_1\}$. $H$ has a subgraph isomorphic to the complete graph $K_2$ on two vertices (actually, one for each ${\Bbb Z}_2$ slice), and since $G$ must have one too, we know $a_1 = -a_0$. Since $S$ generates $\Bbb Z$, $S = \{1, -1\}$. But this $G$ is clearly not isomorphic to $H$; for example, all edges of $G$ belong to some copy of $K_2$, which is not true for $H$.

So there is no such $S$.