Does the number of elements of order $r$ equal $\sum_{|x| = r} |x^G|$?

Question

I want to know if this sentence is true or not?

Let $m_r$ be the number of elements of order $r$ in a finite group $G$ and also let $x$ be an element of order $r$ in this group, then $m_r=\sum_{|x|=r}|x^G|$

where $|x|$ is the order of $x$ in $G$ and $x^G$ is the conjugacy class of $x$.

Answer 1

$$G=A_4\;,\;\;r=2\;,\;\;m_3=3$$

The elements of order $\;2\;$ here are $\;(12)(34)\,,\,\,(13)(24)\,,\,\,(14)(23)\;$ , and the conjugacy class of each of them is three as these elements, together with the unit element, for a normal subgroup of order four in $\;A_4\;$ , but then

$$\sum_{|x|=2}|x^G|=3+3+3=9\neq3=m_2$$

If $\;x^G\;$ means the normal closure the inequality above is, of course, even worse.