Number of particular group conjugation equivalence classes Let $G$ be a finite group and fix $x\in G\setminus\{e\}$, and define a relation (I have shown that this is an equivalence relation) $\sim$ on $G$ by $g\sim h$ if $g=x^{-n}hx^n$ for some $n\in\mathbb Z$. My question is:

How many equivalence classes are in $G/\sim$.

I am having trouble answering this question since it doesn't seem like the equivalence classes will be of the same size. I have examined a similar relation: saying $g$ and $h$ are related if $g=x^nh$ for some $n\in\mathbb Z$, and it is clear that the number of elements in each equivalence class here is $|x|$, so the total number of equivalence classes is $|G|/|x|$. My motivation for the above question is I would like to compare the number of equivalence classes to $|G|/|x|$ (the number of equivalence classes in the other relation discussed in the previous sentence). 

I would be happy with knowing that the number of equivalence classes in $G/\sim$ is greater than $|G|/|x|$, if that is true.

Any help is appreciated.

Alas! Some banging my head on the desk lead me to an answer to my second highlighted problem. It is true that the number of equivalence classes of $G/\sim$ is greater than $|G|/|x|$. This is because, for each $S\in G/\sim$, we clearly have $|S|\leq |x|$, and if $|S_0|$ is the equivalence class containing $x$, then $S_0=\{x\}$, so $|S_0|=1<|x|$. Therefore, the sizes of the equivalence classes in $G/\sim$ are at most $|x|$ with at least one equivalence class having size strictly less than $|x|$, so the total number of equivalence classes must be greater than $|G|/|x|$. 
I will still leave this open, as I think the first highlighted problem is still interesting.
 A: We can count the number of orbits using the centralizers of powers of $x$. And I think we get a sum expression using the Möbius function.
The elements of $C_G(x)$ will all have orbits under $\sim$ of size $1$, since they commute with $x$.
If $d$ is a divisor of $|x|$, then the elements that are in $C_G(x^d)$, but not in $C_G(x^{\ell})$ for some $\ell|d$, will have orbits of size $d$. 
So, for example, suppose that $|x|=p$, a prime. Then the number of orbits is:


*

*One orbit for each element in $C_G(x)$.

*One orbit of length $p$ for each element that is in $C_G(x^p) = G$, but not in $C_G(x)$, each of them counted $p$ times. Thus, we have
$$\frac{1}{p}\left( |C_G(x^p)| - |C_G(x)|\right).$$
If we had $|x|=pq$, with $p\neq q$ both prime, we would have:


*

*One orbit of size one for each element in $C_G(x)$.

*One orbit of size $p$ for each element in $C_G(x^p)$ that is not in $C_G(x)$, counted $p$ times.

*One orbit of size $q$ for each element in $C_G(x^q)$ that is not in $C_G(x)$, counted $q$ times.

*One orbit of size $pq$ for each element in $C_G(x^{pq})$ that is not in $C_G(x^p)$, $C_G(x^q)$, or $C_G(x)$, counted $pq$ times.


So first you add in $C_G(x)$. Then you add $\frac{1}{p}(|C_G(x^p)|-|C_G(x)|)$. Then you add $\frac{1}{q}(|C_G(x^q)|-|C_G(x)|)$. Then you add $\frac{1}{pq}(|C_G(x^{pq}) - (|C_G(x^p)| + |C_G(x^q)|) + |C_G(x)|)$.
(You add back $|C_G(x)|$, because you “took it away” twice, once when you subtracted $|C_G(x^p)|$ and again when you subtracted $|C_G(x^q)|$. 
If I remember my Möbius function correctly, one can express this as a single sum over all divisors of $|x|=n$ by using the Möbius function.
If $\mu(n)$ is the Möbius function, given by $\mu(n) = 1$ if $n$ is squarefree and has an even number of prime factors; $\mu(n)=-1$ if $n$ is squarefree and has an odd number of prime factors; and $\mu(n) = 0$ if $n$ is divisible by the square of a prime. 
Then what we have is (perhaps) equal to something along the lines of
$$\sum_{d|n}\left(\sum_{k|d}\frac{1}{k}\mu\left(\frac{d}{k}\right)\right)|C_G(x^d)|$$
or something like that...
