Show that $G$ is a complete or empty graph on n vertices if and only if every transposition of $\{1, 2, . . . , n\} $ belongs to $Aut(G)$.

Question

Here $Aut(G)$ is the set of all Automorphisms of the graph $G$. One direction of the proof is easy as for all empty or complete graph, every permutation of the vertex set will belong to the set of Automorphisms. However I can't see why the converse is true. Please help me out or point out how to proceed as I am very new to this area.

Answer 1

Suppose that the automorphism group of $G$ contains all transpositions of vertices. Then it contains all permutations of vertices (since the set of transpositions generates $S_n$).

If $G$ does not have any edges, then $G$ is $E_n$ and we are done. Otherwise, $G$ has an edge between two vertices $u$ and $v$. Then $G$ also has an edge between all other pairs $(x,y)$ of (distinct) vertices: take any automorphism which maps $u$ to $x$ and $v$ to $y$ to see that there must be an edge between $x$ and $y$. So in this case, $G$ must be $K_n$.