= Solution
The <conjugate group elements> relation is $x\sim y$ when $y=gxg^{-1}$ for some $g\in G$. It is reflexive by taking the identity, symmetric because $x=g^{-1}yg$, and transitive because $y=gxg^{-1}$ and $z=hyh^{-1}$ imply $z=(hg)x(hg)^{-1}$. Hence it is an <equivalence relation>.
For a finite <group>, the <centralizer> $C_G(x)=\{g:gx=xg\}$ is a subgroup. The map $g\mapsto gxg^{-1}$ has equal values at $g,h$ exactly when $h^{-1}g\in C_G(x)$. Its fibers are therefore the left cosets of $C_G(x)$, giving
$$
\boxed{|\operatorname{Cl}(x)|=[G:C_G(x)]=\frac{|G|}{|C_G(x)|}}.
$$
By <Lagrange's theorem>, this index divides $|G|$. This is also the <orbit-stabilizer theorem> for the conjugation action, with the centralizer as stabilizer.
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