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

Question

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 $I$.

Answer 1

Let $\,\,\{1\}\neq N\triangleleft G\,\,$ , 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

Answer 2

Hint: If $x$ is a fixed point of $N$, then what is $\sigma x$ in relation to $\sigma N \sigma^{-1}$ for some $\sigma\in G$?

Now use $N$'s normality and the transitivity of the $G$-action...