In $G = \operatorname{PGL}_{4}(\mathbb{C})$, a subgroup $E = <a,b,c,d>$ where $$a =\Delta (\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix})_{2}, b =\Delta (\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix})_{2},$$ $$c =\begin{pmatrix} I_{2} & 0\\ 0 & -I_{2} \end{pmatrix},d =\begin{pmatrix} 0 & I_{2}\\ I_{2} & 0 \end{pmatrix}.$$ Here $\Delta(M)_{n}$ means $n$-copies of a matrix $M$ embedded diagonally.
$C_{G}(E)=E$ and $N_{G}(E)/C_{G}(E)=\operatorname{Sp}(4,2).$
Take $A=<a,c>$. I wish to show $\operatorname{GL}(2,2)\leqslant N_{G}(A)/C_{G}(A).$ (It's in fact $\cong$)
Proof. $N_{G}(E)/C_{G}(E)=\operatorname{Sp}(4,2)$ has a subgroup $\operatorname{GL}(2,2)$ stabilizing $A$. Let $g_{i}$ be the generators of $\operatorname{GL}(2,2)$ for $i = 1 ... k$. Thus $$ \cup_{i=1}^{k} C_{G}(E)g_{i} \subseteq N_{G}(A).$$ Because $\operatorname{GL}(2,2)$ acts faithfully on $A$, none of $C_{G}(E)g_{i}$ is in $C_{G}(A).$ Thus, $$ \cup_{i=1}^{k} C_{G}(A)g_{i} \subseteq N_{G}(A)$$ and $\operatorname{GL}(2,2)\leqslant N_{G}(A)/C_{G}(A).$
$\Box$
I realized that the last inclusion doesn't guarantee the implication of the conclusion as the $g_{i}$'s might satisfy some additional relations because $C_{G}(E)\leqslant C_{G}(A)$. Not sure how to fine-tune this..