What are the self-normalizing subgroups of $S_n$?

If we fix $S=S_n=Sym(\Omega)$ where $|\Omega|=n$ then one sufficient condition is that $G=S_\alpha$ is the stabiliser of some $\alpha\in\Omega$.

$g\in N_{S_n}(G)$ with $n\ge 3$ then $S_\alpha=g^{-1}S_\alpha g=S_{\alpha^g}$. $S_\alpha=S_\beta$ if and only if $\alpha=\beta$, so $\alpha=\alpha^g$ (where $S_n$ is acting on $\Omega$ on the right). That is $g\in G$.

By a similiar argument a more general sufficient condition is that $G=S_\Delta$ is the setwise stabiliser of some set $\Delta<\Omega$ with $|\Delta|<\frac{n}{2}$.

Taking this argument further we may consider any partition $\{\Delta_1,\ldots,\Delta_m\}$ of $\Omega$ for which $|\Delta_i|=|\Delta_j|\Rightarrow i=j$, then a sufficient condition is that $G=S_{\Delta_1,\ldots,\Delta_m}$ is the set of permutations which fix each $\Delta_i$.

To see why this is true suppose $g\in N_{S_n}(G)$, then $S_{\Delta_1,\ldots,\Delta_m}=g^{-1}S_{\Delta_1,\ldots,\Delta_m}g=S_{\Delta_1^g,\ldots,\Delta_m^g}$. For each $i$, $|\Delta_i^g|=|\Delta_i|$ and the only orbit of $G$ of size $|\Delta_i|$ is $\Delta_i$ so $\Delta_i^g=\Delta_i$. Hence $g\in G$.