Show that no asymmetric graph $G$ exists with $1 < \big|V(G)\big| \leq 5.$ 
Show that no asymmetric graph $G$ exists with $$1 < \big|V(G)\big| \leq 5\,.$$ 

I tried listing all the possibilities for $\big|V(G)\big| \leq 5$ to prove this statement. I did all for $2$ and $3$, and then it’s getting complicated. Is there any elegant, simple solution to show all $G$ of five vertices or less must have some automorphism other than identity mapping?
 A: It is not hard to prove that a simple graph (on more than one vertex) in which all vertices have degree at most $2$ has a nontrivial automorphism. So any counterexample must contain the following subgraph:

This immediately deals with all simple graphs $G$ with $|V(G)|\leq3$.
Every disconnected graph has automorphisms from its connected components. So it suffices to prove that every connected simple graph with $4\leq|V(G)|\leq5$ has a nontrivial automorphism.
A graph has a nontrivial automorphism if and only if its complement does. So you can restrict to graphs with connected complements. In particular, to graphs $G$ that have no vertex of degree $|V(G)|-1$, so any counterexample must contain the following subgraph:

where $A$ and $B$ cannot share an edge. This leaves only a few graphs $G$ with $|V(G)|=5$. More specifically, those with at least one vertex of degree $3$ and no vertices of degree $4$. There are not many such graphs, you can check this yourself. In fact there are only $5$ such graphs.
