If $G\subseteq S_n$ is a subgroup acting transitively on $\{1,\ldots,n\}$, then a nontrivial normal subgroup $N\subseteq G$ has no fixed points Let $G$ be a subgroup of $S_n$, which acts transitively on $I= \\{1, \ldots, n \\}$. Let $N$ be a nontrivial normal subgroup of $G$. Then $N$ has no fixed points in $I$.

Let $\,\,\\{1\\}\
eq N\triangleleft G\,\,$ , and wlog let us assume $\,\,1\in\\{1,2,...,n\\}\,$ is a fixed point of $\,N\,$. But then for any $\,g\in G\,\,,\,x\in N$ , we get $$g^{-1}xg(1)=1\Longrightarrow xg(1)=g(1)\Longrightarrow g(1)\,\,\text{is a fixed point of } N$$

Well, now use that $\,G\,$ is transitive to get a contradiction