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 $\phi(\operatorname{diag}(1,-1))$.