 Solution

The orbit of $x$ under conjugation is its conjugacy classIt contains $x$ by using $g=e$, so the conjugacy classes cover $G$. If $z=gx{g}^{-1}$, then $x=g^{-1}zg$, and the group action identity shows that conjugating either element further gives the same orbit; hence $C(z)=C(x)$. If $C(x)$ and $C(y)$ intersect at $z$, this gives $C(x)=C(z)=C(y)$. Thus distinct conjugacy classes are disjoint and the conjugacy classes form a partition of a set $G$.