I believe [N:C] = |Aut(H)| always (for H a cyclic subgroup of the symmetric group).
If H is generated by a single cycle, then it is true: whenever H is a regular permutation group, then the normalizer contains the automorphism group fairly literally as a permutation group.
If H is generated by a product of disjoint cycles, then the product of the normalizers of the cycles contains the automorphism group too.
For instance, the normalizer of $H = \langle (1,2,3,4)(5,6,7,8)(9,10)(11,12) \rangle$ contains the (orbit) normalizers $\langle (1,2,3,4), (1,3) \rangle$, $\langle (5,6,7,8), (5,7) \rangle$, $\langle (9,10) \rangle$, and $\langle (11,12) \rangle$. In particular, the normalizer of H induces every automorphism of H, just by working on each orbit.
Proposition Let $G$ be a group of odd order with less than $200$ elements and let $23$ divide the order of $G$. Then $G \cong C_{23}, C_{69}, C_{115}$ or $C_{161}$.
Proof By Cauchy’s Theorem we can find an element $h \in G$ of order $23$. Put $H=\langle h \rangle$. Owing to the $N/C-Theorem$, $N_G(H)/C_G(H)$ can be homomorphically embedded in Aut$(H) \cong C_{23}^* \cong C_{22}$. Since $|G|$ is odd, it follows that $|N_G(H)/C_G(H)|$ must divide $11$. If $|N_G(H)/C_G(H)|=11$, then, since $H \subseteq C_G(H)$, we would have $$|G| \geq\text{ index }[N_G(H):C_G(H)] \cdot \text {index}[C_G(H):H] \cdot |H| \geq 11\cdot 23 = 253,$$contradicting the fact that the order of $G$ is smaller than $200$. We conclude that $ N_G(H)=C_G(H)$.
Observe that $23^2=529 \nmid |G|$, hence $H \in Syl_{23}(G)$. Assume that $\#Syl_{23}(G) \gt 1$. Then by Sylow’s Theorems, we would have $\#Syl_{23}(G)=\text {index}[G:N_G(H)] \geq 24$. But then $|G| \geq 23 \cdot 24 = 552$, again contradicting $G$ having less than $200$ elements. We conclude that index$[G:N_G(H)]=1$, so $G=N_G(H)=C_G(H)$, meaning $H \subseteq Z(G)$.
Finally observe that $|G/Z(G)| \mid |G/H|$. Since $|G|$ is odd, less than $200$ and $|H|=23$, $|G/Z(G)| \in \{1,3,5,7\}$. This means that $G/Z(G)$ is cyclic, and it is well known that in this case $G$ must be abelian. The proposition now follows easily.
Best Answer
That is because you have a homomorphism \begin{align*} N(H)&\longrightarrow \operatorname{Aut}H\\ g&\longmapsto (h\mapsto ghg^{-1}) \end{align*} and the kernel of this homomorphism is the centraliser of $H$.