@Wei Zhou: By the sizes of conjugacy classes of PGL the size of the conjugacy class of an order-k cyclic subgroup $C$ must be $q(q-1)$ or $q(q+1)$.

Now I know what is the size of the conjugacy class of an order-k cyclic subgroup C, but still I don't know what is the number of conjugacy classes of such subgroups C. By Tom's answer it must be $\phi (k)/2$. But why? Can anybody help me?

@Tom: I don't know you how get the number of elements of order k.

@Michael Zieve: Thank you. If possible give me more details of your answer.

@Geoff Robinson: Thank you very much, it was most helpful!

@Geoff Robinson: Thanks. By Derek's answer if $S\unlhd G$ with $S$ simple and $G\leq Aut(S)$, then $p$ is prime divisor of $S$. Whether by your answer it implies that $p$ is prime divisor of $S$?

@Arturo Magindin: Thank you so much for your answer.

@Arturo Magidin: Thanks. Note that $G$ is not $p$-group.