Which elements of a group have equal centralizer? Can we say elements which has equal centralizers are the elements which are in the same coset of $G/Z(G)$? I mean if $C_G(x)=C_G(y)$, then $x=yz$ where $z\in Z(G)$ or $x=y^{-1}$. Are there any other elements of the group which have equal centralizer?
 A: If the centralizer of $x$ is normal, then the other elements in the same conjugacy class as $x$ will have the same centralizer. You can prove this by hand, but its easiest to see with group actions.
A: What is the centralizer of $(1,2,3)$ in $S_5$? Can you think of other elements with the same centralizer? What is the center of $S_5$?
A: The centralizer of a subset only depends on the subgroup generated by that subset. Thus if $n$ is coprime to $|x|$, then $C_G(x) = C_G(x^n)$. 
More generally, $C(x) = C(y)$ iff $C(C(x)) = C(C(y))$ iff $x$ and $y$ are in the same coset of $C(C(x)) \cap C(C(y))$, i.e. $x^{-1}y \in C(C(x)) \cap C(C(y))$. It is an interesting (although not very group-theoretic) question to ask when $C(H) = C(K)$ for two subgroups $H$ and $K$.
A: if $C_G(x)=C_G(y)$ then for any element $a\in H$,
$C_G(x)\leq C_G(a)$ where $H=<x,y>$ is the subgroup generated $x$ and $y$ 
and as a result $C_G(x)=C_G(H)$ as $ C_G(H)=\bigcap C_G(a)$.
proof: Let $a\in H$ then $a=x^{r_1}y^{k_1}...x^{r_n}y^{k_n}$ and $m\in C_G(x)$.
Then clearly $mx^{r_i}m^{-1}=x^{r_i}$ and $my^{k_i}m^{-1}=y^{r_i}$,
$$mam^{-1}=mx^{r_1}y^{k_1}...x^{r_n}y^{k_n}m^{-1}$$
$$mam^{-1}=mx^{r_1}m^{-1}my^{k_1}m^{-1}...m^{-1}mx^{r_n}y^{k_n}m^{-1}$$
$$ =x^{r_1}y^{k_1}...x^{r_n}y^{k_n}=a$$
So $m\in C_G(a)$ so $C_G(H)=C_G(x)$.
