In order to prove this, I did the following:
first, I showed that conjugacy forms an equivalence relation, then I can find its conjugacy classes. I understand how to form a conjugacy class, given a group. So, the index of the normalizer of $\{x\}$ in $G$ is the number of cosets of $N_x$ in $G$, right?
The set $N_x$ should be $\{a: ax = xa\}$, right? How does this set look? I need to find it, so I can take its quotient with $G$ and count how many of them exist. But how to do it? I'm lost.
UPDATE:
I think I got it:
$a\in N_G(H)b \iff ab^{-1} \in N_G(H) \iff ab^{-1}H = Hab^{-1} \iff b^{-1}H = a^{-1}Hab^{-1} \iff b^{-1}Hb = a^{-1}Ha$
So, two conjugates are equal $\iff $ their elements are in the same coset of $N_G(H)$
Thus, there is an explicit bijection between the set of different conjugates and the normalizer, so there are $[G:N_G(H)]$ different conjugates of $H$.