The complete graph on $n$ vertices has exactly one edge joining each pair of distinct vertices, and is denoted by $K_n$
A symmetry of a graph $\Gamma$ is a bijection $\alpha$ taking vertices to vertices and edges to edges such that if $Ends(e) = \{v,w\}$, then $Ends(\alpha(e)) = \{\alpha (v),\alpha(v)\}$. The symmetry group of $\Gamma$ is the collection of all its symmetries. We note this group by $Sym(\Gamma)$
I am trying to show that $Sym(K_n) \cong S_n$. Here's what I have.
I think it is clear that $|Sym (K_n)| \le n!$, because there are $n$ ways to move the first vertex, $n-1$ ways to move the second, etc. (a priori this is an upper bound because I am only looking at the ways of moving the vertices). Thus, I just need to show that there is an injective homomorphism from $S_n$ to $Sym(K_n)$. Let $\sigma \in S_n$. Define $\tilde{\sigma} : K_n \to K_n$ to be $\tilde{\sigma}(v_i) = v_{\sigma (i)}$ and $\tilde{\sigma}(e_i) = e_{\sigma (i)}$, where $v_i$ is a vertex, $e_i$ an edge. it is easy to show that $\tilde{\sigma}$ is a bijection, but showing that it preserves adjacency is a little tougher. I need to show that $Ends(e) = \{v_i,v_j\}$ implies $Ends(\tilde{\sigma}(e)) = \{\tilde{\sigma}(v_i),\tilde{\sigma}(v_j)\}$.
I think uniqueness of edge might help, but I can't figure it out at the moment.