What is the size of $\text {Aut}\ (C)\ $? 
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.
 A: 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.
