I have tried it in the following way.
Let $g\in N(S_{n-1})$.
$(1,2,....,n-1)$ is in $S_{n-1}$.
Now $$g(1,2,....,n-1)g^{-1} = (g(1),g(2),....,g(n-1))\in S_{n-1}$$ Therefore, $g(1)$, $g(2)$, $\ldots$, $g(n-1)$ none of them can be $n$, i.e $g$ fixes $n$.
Therefore $g\in S_{n-1}$.
Hence normaliser of $S_{n-1}$ in $S_n$ is $S_{n-1}$.
Is it correct? Is there any other way to find it?