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.