Intersection of $PGL_2(q_0)$’s in $PGL_2(q_0^3)$

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 integer at least $2$. 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 $Lcong C_{q_0-1}$, then $|x(L)|=q_0^2+q_0 = |H: L|$.
  2. If $Lcong C_{q_0+1}$, then $|x(L)|=q_0^2-q_0 = |H:L|$.
  3. If $Lcong (q_0)$, then $|x(L)|=q_0^2-1 = |H:L|$.

Note that if $Lcong {1}$, 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 $aneq 3$. 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 $|N_G(L)|/|N_H(L)|$ and 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.