I am attempting to solve this problem given to me. I know that the Folkman graph can be constructed in such a way:
$V_1 \mapsto V_2$, where $V_1$ is a set where exactly two vertices have the same neighbourhood.
Now, we need to show that $G$ is vertex intransitive. Is it because if we map $a$ to one of its neighbours $b$, then such a graph permutation would require some vertex in $V_1$ to map to a vertex in $V_2$ it isn't adjacent to? Any help with proving vertex-intransitivity would be much appreciated.
The vertices in $V_1$ can be distinguished from the vertices in $V_2$ by the following property: for each $v \in V_1$, there is another vertex $v'$ with the same neighborhood.
This property is preserved by graph isomorphism, and therefore any automorphism of $G$ must map $V_1$ to $V_1$ and $V_2$ to $V_2$.