D

连接$CH$并延长交$AD$于$P$，连接$PE$，

$\because$四边形$ABCD$是正方形，
$\therefore \angle A=90^{\circ}$，$AD$∥$BC$，$AB=AD=BC=4\sqrt{2}$，
$\because E$，$F$分别是边$AB$，$BC$的中点，
$\therefore AE=CF=\frac{1}{2}×4\sqrt{2}=2\sqrt{2}$，
$\because AD$∥$BC$，
$\therefore \angle DPH=\angle FCH$，
在$\triangle PDH$和$\triangle CFH$中，
$\left\{\begin{array}{l}{∠DPH=∠FCH}\\{∠DHP=∠FHC}\\{PH=FH}\end{array}\right.$，
$\therefore \triangle PDH$≌$\triangle CFH\left(AAS\right)$，
$\therefore PD=CF=2\sqrt{2}$，
$\therefore AP=AD-PD=2\sqrt{2}$，
$\therefore PE=\sqrt{A{P}^{2}+A{E}^{2}}=4$，
$\because$点$G$，$H$分别是$EC$，$FD$的中点，
$\therefore GH=\frac{1}{2}EP=2$.
故选：$D$.