There are $8$ conjugates of $H$ in $G$, where $H$ is a Sylow $p$-subgroup with $p=7$

Question

Let $H$ be a subgroup of order $7$ in group $G:=PGL(2,7)$ (of order $336$). Show that there are $8$ conjugates of $H$ in $G$.

I know that $H$ is a Sylow $7$-subgroup so let $n_7$ denote the number of Sylow $7$-subgroups in $G$. Since $n_7$ must be $1$ mod $7$ and must divide $48 = \frac{|G|}{7}$ then $n_7$ is either $1$ or $8$. I am not sure how to rule out $n_7 = 1$.

Thanks for any help!