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The normalizer of $G < S_n$ in the group $S_n$ is defined as $$N_{S_n}(G) = \{\pi \in S_n \mid \pi G = G \pi \}.$$ The group $G$ is self-normalizing in $S_n$ iff $$N_{S_n}(G) = G.$$

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

This is equivalent to asking: Given a subgroup $G < S_n$, under what condition it is not a normal subgroup of any other $H \leq S_n, G\neq H$?

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    A neater way of writing this is 'For which $G$N(G)=G$ where $N(G)$ is the normaliser of $G\in S_n$'. Also in your implication you want $G$G=H$ – 2017-01-13
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    Great suggestion! – 2017-01-13
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    This is a bit like giving the answer to a problem and asking for the problem. I don't believe that this question has a sensible (i.e. non-tautological) answer in that degree of generality, because $S_n$ has a very large number of subgroups, some of which are self-normalizing and others not. You probably have some specific types of conditions in mind, so you should try asking something more specific. – 2017-01-13
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    I'm looking for sufficient conditions for this. Is there any property of $G$ that ensures it is self-normalizing? – 2017-01-13
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    Right, but that's rather trivial. Anything non-trivial? Here https://groupprops.subwiki.org/wiki/Free_factor_implies_self-normalizing_or_trivial it says that free factorization may give a clue. That is we are looking for non-trivial free factors of $S_n$. The question becomes, how do we get these factors? – 2017-01-13
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    Any group with a nontrivial free factor is infinite. – 2017-01-13

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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 $g\in G$.

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    Thanks! This seems to cover a broad class of subgroups. I get the main point. Could you please clarify some points in your answer: 1) Why $n \geq 3$? 2) why $|\Delta| < \frac{n}{2}$? and most importantly could you elaborate on the last statements that give the most general form of $G$? How did you get that? – 2017-01-14
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    If $|\Delta| = n/2$ then it is not self-normalizing. – 2017-01-14
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    notice that if $n=2$ then $S_\alpha=S_\Delta$ with $|\Delta|=n/2$ and notice that if $|\Delta|=n/2$ then $S_\Delta=S_\Delta,\Omega-\Delta$ so we have a partition $\{\Delta_1,\Delta_2\}$ with $|\Delta_1|=|\Delta_2|$. Therefore all your questions reduce to understanding the largest class of groups I've given. I'll edit the answer to give an explanation of it – 2017-01-14