Let $(G,\cdot)$ be a centerless group (i.e. $Z(G)=\{e\}$) with $n$ elements, where $n$ is an integer which is not divisible by $4$. Is it true then that $\exists a,b \in G$, $a\ne b$ such that $C(a)=C(b)$?
I tried to use the class equation, but I wrote it wrongly and as a result I made no progress. Here is basically what I tried to do and it is definitely wrong.
I don't really know what else to try. I thought about assuming the contrary and trying to reach a contradiction, but I got stuck really quick.
Given that $C(a)$ means "centralizer of $a$ [in $G$]":
Since $G$ is centerless, it is either trivial or nonabelian. If it is trivial, you have no hope, so I guess it's supposed to be non-trivial. The order cannot be a prime power, so $|G|$ must be divisible by an odd prime $p$. Let $a$ be an element of order $p$. Let $b=a^{-1}$.
Note that the hypothesis on $|G|$ is irrelevant: if $G$ is centerless, it cannot be a nontrivial $p$-group, so you know for sure $|G|$ is either $1$ or divisible by an odd prime, and the result follows.
But the result also follows for non-centerless groups: if $Z(G)$ is nontrivial, pick $g\in Z(G)$, $g\neq e$; then $C(g)=C(e)=G$. A better question then would be whether you can find such elements with $C(a)=C(b)\neq G$.
The only obstacle to the same answer would be if for every $g\in G$, either $g\in Z(G)$ or $g=g^{-1}$. I claim that this is impossible.
Indeed, if you had such a group $G$, then $G/Z(G)$ is abelian, so $G$ is nilpotent of class $2$ and is the product of its $p$-parts. The odd $p$ parts are all abelian, so we are reduced to a $2$-group of class exactly $2$ with every noncentral element of order $2$. Let $x$ be noncentral and $y$ be an element of $G$ that does not commute with $x$. Then $\langle x,y\rangle$ is a nonabelian group of class $2$, generated by two elements of order $2$. The only possibility is that it is the dihedral group of order $8$, but that group has a noncentral element of order $4$, and hence that element would be noncentral in $G$ of order $4$, a contradiction. So if $G$ is not abelian, then you can always find an $a$, $a\neq a^{-1}$ with $C(a)\neq G$.