A Cayley graph is a vertex-transitive graph, meaning that given any two vertices $v_1, v_2$ in the vertex set of the graph $X$ ($V(X)$), there exists an automorphism $\phi:V(X) \rightarrow V(X)$, such that $\phi(v_1)=v_2$.
For a Cayley graph say, $X=Cay((\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_q, \{s,t\})$, where $|s|=q,|t|=p$, let $\theta: \mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$ be known.
As an example, when $q=3$, $\theta_s(t^iu^j)=s^{-1}t^iu^js=(t^{-j}u^{i-j})$, where ${t,u}$ forms a basis of $(\mathbb{Z}_p \times \mathbb{Z}_p)$ with $u= \theta_s(t)$.
For a vertex $v_1=(t^{k_1}u^{k_2},s^{k_3})$, if I trace along a sequence of edges, say $sts^{-1}t^2$, then it ends up at the vertex $v_2=(t^{m_1}u^{m_2},s^{m_3})$ (i.e. $v_1 sts^{-1}t^2 =v_2$).
Is it possible to find an automorphism $\phi:V(X) \rightarrow V(X)$, which will map $v_1$ to $v_2$ if I know the values of $k_1, k_2, k_3, m_1, m_2, m_3, t, s, \theta_s, p, q$ such that it will satisfy the following condition?
If I calculate the $\phi(v^\prime)$ for any other vertex $v^\prime \in X$, and get $\phi(v^\prime)=w$, then $w$ has to be a vertex present at a point obtained by tracing along the same sequence of edges (path) $sts^{-1}t^2$ starting at $v^\prime$. i.e. $v^\prime sts^{-1}t^2 =w$.
How to get such an automorphism if possible?
Thanks a lot in advance.