Suppose $G$ is a group that is transitive on $\{1,2,\dots,n\}$, and let $K_i=\{g\in G \;|\; g(k)=k \;\forall k=1,2,\dots,i\}$ for $i=1,2\dots,n$.
Prove that $G=S_n$ iff $K_i\neq K_j$ for all pairs $i,j$ such that $i\neq j$ and $i<n-1$.
$(\implies)$ Assume $i<j$. Take $g=(j\; j+1)$. Clearly $g\in K_i$ but $g\not\in K_j$. Hence $K_i\neq K_j$.
$(\impliedby)$ Note that $$K_n=K_{n-1}<\dots< K_2 < K_1$$
Hence $n \mid |G|\mid n!$. Write $|G|=kn$ where $k$ divides $(n-1)!$.
By Orbit-Stabilizer Theorem, $|K_1|=k$.
I need idea to prove that $k=(n-1)!$