TITLE: Normalizer of the cyclic group in $S_n$
QUESTION [9 upvotes]: Let $G = S_n$ and $H = \langle (1,2,\ldots,n) \rangle.$ It is not too hard to see that $$C_G(H) = H.$$ 
What I am now wondering is, which group is $N_G(H)?$ Is there any way to determine that? 
I guess the key is to use the fact that $$g (1,2,\ldots, n) g^{-1} = (g(1), \ldots, g(n))$$ but I don't see how to get the full answer just by this.

REPLY [5 votes]: Perhaps it is worth noting the more general fact, that if $H$ is a group of order $n$, and $\psi : H \to S_{n}$ is the regular representation, then $N_{S_{n}}(\psi(H))$ is isomorphic to the holomorph $H \rtimes \operatorname{Aut}(H)$ of $H$.