In $G=\operatorname{PGL}(4,5)$ there are two elementary abelian $2$-subgroups of order $16$ denoted by $E\_{1}$ and $E\_{2}$ with $N\_{G}(E\_{1})=E\_{1}.\operatorname{Sp}(4,2)$ and $N\_{G}(E\_{2})=E\_{2}.(2^{3}:S\_{3})$.


$N\_{G}(E\_{1})$ is a maximal subgroup of $G$ and $(2^{3}:S\_{3})$ is a point stabilizer of a nonidentity element of $E\_{2}$ in $\operatorname{Sp}(4,2)$ (viewing $E\_{2}$ as a symplectic basis of $\operatorname{Sp}(4,2)$).


By comparison, $H=\operatorname{PGL}(4,3)$ has $N\_{H}(F\_{1})=F\_{1}.M\_{1}$ and $N\_{H}(F\_{2})=F\_{2}.M\_{2}$ where $F\_{i}$ are again elementary abelian $2$-subgroups of order $16$ and $M\_{i}$ are, by some Magma computation, two maximal subgroups of $\operatorname{Sp}(4,2)$ of orders $72$ and $120$, respectively. I see that $2^{4}.\operatorname{Sp}(4,2)$ is not a subgroup of $H$ any more, in contrast. (It appears in the corresponding finite unitary group now.) Also $M\_{1}$ and $M\_{2}$ are stabilizers in $\operatorname{Sp}(4,2)$ of $f\_{1}\in F\_{1}$ and $f\_{2}\in F\_{2}$ which have orbit sizes $10$ and $6$.


Is there an explanation for this phenomenon? What are these orbits of sizes $10$ and $6$ now? I hope to generalizer this. For instance, in $S=\operatorname{PGL}(8,3)$, what are the analogous orbit sizes in $2^6$?


I've only just found out that $M\_{1}\cong \operatorname{SO}^{+}\_{4}(2)$ and $M\_{2}\cong \operatorname{SO}^{-}\_{4}(2)$. They do have orbit sizes 1,5,10 and 1,9,6….



Normalisers and stabilisers in classical groups $\operatorname{PGL}_{4}$
The difference between the examples arises principally because $5 \equiv 1 \bmod 4$ and $3 \equiv 3 \bmod 4$.


For $q \equiv 1 \bmod 4$, $G := {\rm GL}(4,q)$ has centre $Z$ divisible by $4$, and contains a group $S$ of symplectic type with $N\_G(S) = ZS.{\rm Sp}(4,2)$. The group $S$ maps onto your group $E\_1$ in ${\rm PGL}(4,q)$.


As you pointed out, you get similar behaviour in the unitary group when $q \equiv 3 \bmod 4$


When $q \equiv 3 \bmod 4$, $G := {\rm GL}(4,q)$ does not have an element of order $4$ in its centre. It has extraspecial subgroups $S^+$ and $S^-$ of two different types, with normalizers $ZS^+ {\rm GO}^+(4,2)$ and $ZS^- {\rm GO}^+(4,2)$ The groups $S^+$ and $S^-$ map onto your groups $F\_1$ and $F\_2$.


The inverse image of your group $E\_2$ when $q=5$ has centre of order $4$ and derived group of order $2$ (as for $E\_1$) but, unlike the inverse image of $E\_1$, it has exponent $8$ and a smaller normalizer.


