Conjugacy classes of elements of a prime order in $PSL_2(q)$ Let $q=p^f$ be a prime power. Given a prime number $r$, how many conjugacy classes of elements of order $r$ are there in $PSL_2(q)$? This topic should have appeared in literature, and I am told (without any proofs or references) that there is exactly one conjugacy class of involutions in $PSL_2(q)$. If $q\equiv1\pmod{4}$, I think this may be proved as follows:
Denote the quotient map from $SL_2(q)$ to $PSL_2(q)$ by $\phi$. Let $\phi(A)$ be an involution in $PSL_2(q)$ with $A\in SL_2(q)$. Then $A^2=\mu I$ for some $\mu\in\mathbb{F}_q$, and $\mu=\pm1$ since $\det(A)=1$. Let $\lambda_1,\lambda_2$ be the roots of $|\lambda I-A|=0$ in the algebraic closure of $\mathbb{F}_q$. We have $\lambda_1\lambda_2=\det(A)=1$ and $\lambda_1^2=\lambda_2^2=\mu$. Viewing that all the fourth roots of unity lie in $\mathbb{F}_q$, we conclude that $\lambda_1,\lambda_2$ are the eigenvalues of $A$ in $\mathbb{F}_q$. If $\mu=1$, then $\lambda_1=\lambda_2=\pm1$, and there exists $B\in GL_2(q)$ such that
$$
A=B
\begin{pmatrix}
\lambda_1&\nu\\
0&\lambda_2
\end{pmatrix}
B^{-1},
$$
where $\nu\in\mathbb{F}_q$. However, $A^2=I$ implies $\nu=0$, and this leads to $A=\pm I$, which contradicts $o(\overline{A})=2$. Hence we have $\mu=-1$. Let $i$ be the element in $\mathbb{F}_q$ such that $i^2=-1$. Then either $\lambda_1=i=-\lambda_2$ or $\lambda_1=-i=-\lambda_2$, and there exists $C\in SL_2(q)$ such that $A=C\operatorname{diag}(\lambda_1,\lambda_2)C^{-1}$, i.e., $A=\pm iC\operatorname{diag}(1,-1)C^{-1}$. This indicates the assertion by saying that every involution in $PSL_2(q)$ is conjugate to $\phi(\operatorname{diag}(1,-1))$.
 A: There are several questions here, and it depends how much group theory you want to use. To answer the question about involutions, the methods are different depending whether or not $p =2$ ( you did not exclude the possibility $p = r$ as far as I could see). If p =2, then ${\rm PSL}(2,q) = {\rm SL}(2,q).$ The subgroup of diagonal matrices has order $q-1$ and the only scalar matrix it contains is the identity. This group acts by conjugation on the upper triangular unipotent matrices, and each non-identity element fixes only the identity. Hence the action is transitive, and all unipotent unitriangular matrices are conjugate in this case.
By Sylow's theorem, every involution of ${\rm SL}(2,q)$ is conjugate to a unipotent upper triangular matrix in this case.
  Now consider the case that $p$ is odd. If $q =3,$ the result is clear, so suppose that $q > 3.$ Then ${\rm PSL}(2,q)$ is a non-Abelian simple group, in particular a perfect group. Is Sylow $2$-subgroup is either elementary Abelian of order $4$ or a dihedral $2$-group. In the former case, we have $N_{G}(S) > C_{G}(S)$ for $S$ a Sylow $2$-subgroup of $G,$ so that $[N_{G}(S):C_{G}(S)] = 3,$ and all non-identity elements of $S$ are conjugate. In the latter case, it follows as a textbook application o Alperin's fusion theorem and transfer theory that $G$ has only one conjugacy class of involutions.
   When $r =p$ is odd, there are two conjugacy classes of elements of order $p$ in ${\rm PSL}(2,p^{f}),$ by an argument similar to the one above, except that the orbit sizes under conjugation action by (images of) diagonal matrix have length $\frac{q-1}{2}$ because in ${\rm SL}(2,q)$, $-I$ is diagonal.
  If $r$ is odd, but $r \neq p,$ then there are $\frac{r-1}{2}$ conjugacy classes of elements $r$ in 
(I omit details, but a Sylow $r$-subgroup is cyclic, and for $R$ a Sylow $r$-subgroup, we have $[N_{G}(R):C_{G}(R)] \leq 2,$ but the index must be greater than $1$ by Burnside's transfer theorem).
Later edit: Actually, I realise there is a shortcut for involutions. By the Thompson transfer lemma, every involution is conjugate to an involution in a cyclic maximal subgroup of a Sylow $2$-subgroup, but that cyclic maximal subgroup has only one involution. The Thompson transfer lemma states that if $S$ is a finite group of even order with no factor group of order $2,$ and $M$ is a subgroup of $G$ whose index is even, but not divisible by $4,$ then every involution of $G$ is conjugate to an involution of $M$. To outline the proof, let $t$ be an involution f $G$ with conjugate in $M.$ Then $t$ act fixed point freely by right translation on the right cosets of $M$ in $G.$ Since $[G:M]$ is not divisible by $4,$
$t$ acts as a product of an odd number of transpositions in that permutation action of $G$, so as an odd permutation, contrary to the fact that $G$ has no factor group of order $2.$
