Intersection of $\mathrm{PGL}_2(q_0)$'s in $\mathrm{PGL}_2(q_0^3)$

Question

$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}$I would like an explanation for some strange numerology which I encountered when studying intersections of subfield subgroups in $\PGL_2(q)$. Here's the situation...

Let $G=\PGL_2(q)$ and let $H=\PGL_2(q_0)$ with $q=q_0^a$ for $a$ an odd integer at least $3$. Let $H_1$ be a distinct conjugate of $H$ in $G$. I believe that the possible intersections of $H$ and $H_1$ are isomorphic to one of the following: $$ \{1\}, \,\, C_{q_0-1} \,\, C_{q_0+1}\,\, \textrm{or} \,\, [q_0]. $$ Note that, provided $q_0>3$, there is one conjugacy class of groups of each of these sizes in $H$.

Suppose that $L$ is a subgroup of $H$ of one of the above isomorphism types. Define $$ x(L) = \{ H_3 \mid H_3 \in H^g, \, H_3 \cap H=L\}. $$ Now assume that $q=q_0^3$ with $q_0>3$. Under this assumption we have the following remarkable numerology:

1. If $L\cong C_{q_0-1}$, then $|x(L)|=q_0^2+q_0 = |H: L|$.
2. If $L\cong C_{q_0+1}$, then $|x(L)|=q_0^2-q_0 = |H:L|$.
3. If $L\cong [q_0]$, then $|x(L)|=q_0^2-1 = |H:L|$.

Note that if $L\cong \{1\}$, then $|x(L)|=q_0^6-q_0^3-q_0^2+q_0+1$ which is not such an interesting number. One could extend the definition of $L$ to include $\PGL_2(q_0)$ itself, in which case $|x(L)|=1=|H:L|$ but whatever. There does not seem (to me) be an obvious reason for these sets to have size $|H:L|$ in each case. It certainly depends upon the fact that $q=q_0^3$ -- it would not be true if $q=q_0^a$ for $a\neq 3$. Perhaps there is some group isomorphic to $H$ acting on the set $x(L)$?

Some notes:

1. If we want to calculate $|x(L)|$ for $L$ non-trivial, then there is an easy method. We observe that any pair of $G$-conjugates of $L$ in $H$ are, in fact $H$-conjugate. It is then easy enough to prove that the number of conjugates of $H$ that contain $L$ is $|N_G(L)|/|N_H(L)|$. Since the possible intersections are so restrictive, one then obtains that $$|x(L)|=\frac{|N_G(L)|}{|N_H(L)|} - 1. $$ So I am asking why, in all cases when $L$ is non-trivial, we should have $$ \frac{|N_G(L)|}{|N_H(L)|} - 1= |H:L|. $$
2. I ran some GAP code to check one possibility: suppose that $G=\PGL_2(5^3), H=\PGL_2(5)$ and $L$ is a cyclic subgroup of $H$ of order $4$. Consider the corresponding set $x(L)$, a set of size 30 whose elements are conjugates of $H$. I wondered if this set might be an orbit of some group $H_2$, a conjugate of $H$ in $G$, acting on $H^G$ by conjugation. But computer says NO.
3. If you prefer to think in $\GL_2(q)$ rather than $\PGL_2(q)$, then please crack on.