Consider the graph $C$
How do I find the size of $\text {Aut}\ (C)\ $?
Since each of the vertices has $3$ neighbouring points so $\text {Aut}\ (C)$ acts transitively on the set of all vertices of the graph $C.$ Let us fix some element say $1 \in V_C,$ where $V_C$ is the collection of all vertices of $C.$ Then we find that $\text {Orb}\ (1) = \{1,2,\cdots, 8 \}.$ If we can find out $\text {Stab}\ (1)$ we are through by orbit-stabilizer theorem. How do I find the stabilizer of $1\ $? Any help in this regard will be highly appreciated.
Thanks for reading.
This is not an answer to the question, rather the final nail in the coffin of an assertion made in the question: An example of a graph $G$ such that every vertex has degree three but $Aut(G)$ is not transitive.
Not having any graph-drawing software handy: Say the vertices are $(n,0)$ and $(n,2)$ for $n\in\Bbb Z$ and also $(0,1)$ and $(1,1)$. Note that when I say we add an edge $(p.q)$ of course we also add the edge $(q,p)$. Add edges $((n,j),(n+1,j))$ for $n\in\Bbb Z$ and $j=0,2$. Add edges $((n,0),(n,2))$ for $n\in\Bbb Z\setminus\{0,1\}$. Also edges $((0,0),(0,1)), ((0,1),(0,2)),((1,0),(1,1)), ((1,1),(1,2)), ((0,1),(1,1))$. A bi-infinite ladder with a little extra stuff in the middle of two of the steps.
If I've given the formal definition of the graph I have in mind properly, then the vertex $(0,1)$ has the property that starting there you can traverse exactly five edges and get back to where you started. The vertex $(10,0)$ does not have this property, so $Aut(G)$ is not transitive on vertices.
Exercise Give an example of a finite graph with the same two properties.