Is it true that $C(x)=C(y)$ 
Suppose $G$ is a finite group. Let $x,y\notin Z(G)$, where $Z(G
)$ is centre of $G$.
If $yx=xy$, is it true that $C(x)=C(y)$ where $C(y)$ denotes the centralizer of $y$ in $G$.

My try:
I think the answer is false. But I tried for $G=D_n$ where $D_n$ is dihedral group for which the result is true I guess. But I am unable to prove it or find a suitable counter-example.
Can someone please help me out?
 A: This is false. Consider the symmetric group. Two elements commute in the symmetric group when their cycle decompositions are disjoint. So if $x = (12) $ and $y = (34)$, and if these cycles are in $S_5$, then $x$ commutes with $y$, but their centralizers are different. For example, $(45)$ commutes with the first, but not the second.
A: The original question had no restriction on $x$ ("Let $y\notin Z(G)$, where $Z(G
)$ is centre of $G$. If $yx=xy$, is it true that $C(x)=C(y)$ where $C(y)$ denotes the centralizer of $y$ in $G$."). So assuming that $x$ can be anything, this statement is false always - there is no group where this holds.
As $y\not\in Z(G)$, we have $C(y)\lneq G$. Then taking $x\in Z(G)$ (possibly trivial) we have $yx=xy$ and $C(x)=G$. Therefore, $yx=xy$ but $C(x)\neq C(y)$.

The new question, with $x, y\not\in Z(G)$, can also be seen to be false by constructing a counter-example to the original question where $x$ is non-trivial, and then embedding this example into a centerless group. That is:
Set $H=S_3\times \mathbb{Z}_2$. Then $Z(H)=\mathbb{Z}_2$, so let $x\in Z(H)$ be non-trivial and let $y\in H\setminus Z(H)$. Then $xy=yx$ but $C_H(y)\lneq C_H(x)=H$.
Consider the embedding of $H$ into $G=S_{|H|}$ (got via Cayley's theorem). As $Z(G)$ is trivial, we have that $x, y\not\in Z(G)$. Therefore, we have $x, y\not\in Z(G)$ with $xy=yx$, and as $C_H(x)\neq C_H(y)$ we have $C_G(x)\neq C_G(y)$ as required.
