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Given the normalizer N(G) of a group $G < S_n$, is $G$ uniquely defined? In either case is there a procedure to find "a" $G < S_n$ with given normalizer $N(G)$?

Does such $G$ always EXIST?

The normalizer is defined as $N_{S_n}(G) = \{\pi \in S_n | \pi G \pi^{-1} = G\}$

Added followup questions:

Is there a property that guarantees a given group $H

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    I don't think so: two normal subgroups have the same normaliser, namely $S_N$. – 2017-01-18
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    I am not aware of any method to find $G$ given $N(G)$, but $G$ is certainly not unique. I think the smallest example is in $S_4$ with $N(G)=S_3=\langle (1,2),(1,3)\rangle$. You can check that this is satisfied if $G=S_3$ or if $G=C_3=\langle (1,2,3)\rangle$ – 2017-01-18
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    @Bernard is correct. For most $n$, the only proper non-trivial normal subgroup of $S_n$ is $A_n$, but if $n=4$, then $S_4$ has two proper non-trivial normal subgroups, $A_4$ and $K_4=\langle (1,2)(3,4),(1,3)(2,4)\rangle$ – 2017-01-18
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    Ok, that makes sense. But how about the main part? How do we go backwards? any ideas? – 2017-01-18
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    There is an obvious procedure. Find all normal subgroups $G$ of the given group $N$ and for each such $G$ check whether $N$ is equal to the full normalizer of $G$ in $S_n$. I think the question is too general to say much more, so you would need to make further assumptions if you wanted a more directed answer. – 2017-01-18
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    Right, thanks. We are, assuming checking for the normalizer of a group within $S_n$ has a efficient (polynomial in $n$) solution. Also do we know that such a subgroup exists? – 2017-01-18
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    It is widely believed that there is no polynomial time algorithm for checking whether a given subgroup of $N_{S_n}(G)$ is the full normalizer of $G$ in $S_n$. It may even be NP-complete, although that is also unknown. In both theory and practice, computing normalizers is one of the most difficult problems in algorithmic group theory. That is why I think we need extra assumptions in order to make progress. – 2017-01-18
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    I don't understand your question about does $G$ always exist because you started by assuming that $N(G)$ was the normalizer of $G$. There certainly exist subgroups of $S_n$ that are not equal to the normalizer of some other subgroup. – 2017-01-18
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    Robert's example generalizes to normalizer of $p$-Sylow subgroups in general. – 2017-01-19
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    Thanks for your explanation on the complexity of the normalizer. I'm not sure if I should post this as a separate question, but then I'd like to ask, what general assumption on G, or N(G) allows us to efficiently go from N(G) to $G$? Also is there a property that guarantees a group $G < S_n$ is $N(G)$ for some $G < S_n$? (I can add this to the initial question maybe) – 2017-01-19

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