How to compute the normalizer $N(H)$? 
Let  $H$ be  a   subgroup   generated  by $(12)$  in $S_3$. Compute the  normalizer  $N(H)$ of $H$.

 My attempt:
$\sigma$ $\in N(H) $ then $ \sigma H \sigma^{-1}=H=(12)$
implies $\sigma (12)= (12)\sigma$ 
Now applying  both sides  to $3$
$\sigma (3) (12) =(12) \sigma(3)$
$(12)$ cancel both  each other
$\sigma (3)  = \sigma(3)$
I thinks  I'm  using  wrong logic.
Any  hints/suggestion  on how to compute the normalizer $N(H)$?

Hint: For any $\sigma\in S_n$, we have $$\sigma(12)\sigma^{-1}=(\sigma(1)\sigma(2)),$$ where $\sigma(i)$ is $\sigma$ applied to $i$, with $i$ from the set $S_n$ acts on.
