$G$ finite, the number of distinct conjugates of $x$ is the index of the normalizer $N_x$ of $\{x\}$ in $G$ 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 $H$.
 A: If you're familiar with the orbit stabilizer theorem,  this result is a quick application. (If you aren't familiar with this result, it still might be useful to know that this problem is a special instance of a more general phenomenon!)
The theorem states that if $G$ is a finite group acting on a set $S$, and $O$ is the orbit of some $x \in S$, then 
$$|O| = [G: \operatorname{Stab}_x]. \,\,\,\,\,\, (*)$$
That is, the size of the orbit is the index of the stabilizer.
In your case, consider the action of $G$ on itself by conjugation. Assuming $(*)$, do you see how to finish?
A: What I like to do is to set up an explicit bijection between the set of conjugates of $x$ and the set of cosets of $N_x$. 
Now the set of conjugates of $x$ looks like $ \{y^{-1}xy \ | \ y \in G\}$. The obvious mapping,  
$$ y^{-1}xy \; \mapsto \; \text{right coset of $N_x$ for $y$} $$
works just fine.Show that this map is one-to-one which is a fun little exercise. The fact that it is onto is plain. 
Then you can argue that the domain and the range of the bijective map above must have the same number of elements as long as they are finite. The number of elements in the range of the bijection is the index of $N_x$in $G$. 
