Let $N$ be a normal subgroup of a finite group $G$

Consider $G$ acting on $N$ via conjugation. Every element $g \in G$ defines an homomorphism of groups $N \to N$. Now, it is known that $\operatorname{Aut}(C_5)=C_4$, so the above action defines a map $G \to C_4$.
But, since $|G|$ is odd, every element of $G$ has odd order.
It follows that the homomorphism defined by conjugation has to be the trivial one, so every element of $G$ acts trivially on $N$, that is, for any $x \in N, g \in G$ $$gxg^{-1}=x$$ which implies $$gx = xg$$ so $N \subseteq Z(G)$.

By using N/C Lemma, $G/C_G(N)$ is isomorphic to a subgroup of $Aut(N)$.
Note that $|Aut(N)|=4$.
Since $|G|/|C_G(N)|$ divides $|Aut(N)|$ and $|G|$ is odd, we have $G=C_G(N)$, that is $N\leq Z(G).$