My understanding of this so far is that for $G$ to have a non-trivial automorphism group, ${\rm Aut}(G)$ must be greater than $1$ and therefore there exists a non-trivial symmetry of the vertices of $G$, i.e. not the identity. Hence, $G$ is symmetric (not asymmetric). So to prove all graphs on $4$ vertices have non-trivial automorphism group I can prove that there are no asymmetric graphs on $4$ vertices.
I believe my reasoning is correct but I don't know how to proceed from here. If anyone could help I would really appreciate it.
Best Answer
This is,of course, a matter of a finite check that you can leave to a computer. However, for a "manual" proof, we need a way to classify the (numerous) cases.
First notice that a graph with $4$ nodes cannot have all the degrees of those nodes different. Namely, there is no graph with precisely the degrees $0,1,2,3$, because the node with degree $0$ is not linked to any nodes, while the node with degree $3$ is linked to all nodes, including the former - a contradiction.
Thus, we can distinguish the cases based on the degree of two nodes with the same degree: