C

$\because$四边形$ABCD$和四边形$BGEF$都是正方形，
$\therefore DC=BC=AD=AB$，$FE=FB=BG=EG$，$\angle BCD=\angle BAD=\angle BFE=\angle BGE=90^{\circ}$，
$\therefore DB=\sqrt{A{D}^{2}+A{B}^{2}}=\sqrt{2}AB$，$EB=\sqrt{F{E}^{2}+F{B}^{2}}=\sqrt{2}FB$，$\angle CDB=\angle CBD=\angle ABD=\angle ADB=\angle FBE=\angle FEB=45^{\circ}$，
$\therefore \frac{AB}{DB}=\frac{FB}{EB}=\frac{\sqrt{2}}{2}$，$\angle ABF=\angle DBE=45^{\circ}-\angle DBF$，
$\therefore \triangle ABF$∽$\triangle DBE$，
故①正确；
延长$AF$交$BD$于点$L$，
$\because \angle FAB=\angle EDB=45^{\circ}$，$\angle ABD=45^{\circ}$，
$\therefore \angle ALD=\angle FAB+\angle ABD=90^{\circ}$，
$\therefore AF\bot BD$，
故②正确；
$\because \angle HEB=\angle EDB=45^{\circ}$，$\angle EBH=\angle DBE$，
$\therefore \triangle EBH$∽$\triangle DBE$，
$\therefore \frac{BE}{BD}=\frac{BH}{BE}$，
$\therefore BE^{2}=BH\cdot BD$，
$\therefore BE^{2}=BG^{2}+EG^{2}=2BG^{2}$，
$\therefore 2BG^{2}=BH\cdot BD$，
故③正确；
连接$CG$，设点$E$运动的时间为$t$，
$\because BD=\sqrt{D{C}^{2}+B{C}^{2}}=\sqrt{2}BC$，$BE=\sqrt{E{G}^{2}+B{G}^{2}}=\sqrt{2}BG$，
$\therefore \frac{BC}{BD}=\frac{BG}{BE}=\frac{\sqrt{2}}{2}$，
$\because \angle GBE=\angle GEB=\angle CBD=45^{\circ}$，
$\therefore \angle CBG=\angle DBE=45^{\circ}-\angle CBE$，
$\therefore \triangle CBG$∽$\triangle DBE$，
$\therefore \angle GCB=\angle EDB=45^{\circ}$，$\frac{CG}{DE}=\frac{BC}{BD}=\frac{\sqrt{2}}{2}$，
$\therefore$点$G$由点$C$沿射线$CG$方向运动，$CG=\frac{\sqrt{2}}{2}DE$，
$\because$点$D$的速度为$\frac{DE}{t}=\frac{1}{t}DE$，点$G$的速度为$\frac{\frac{\sqrt{2}}{2}DE}{\;t}=\frac{\sqrt{2}}{2t}DE$，
$\therefore \frac{\frac{\sqrt{2}}{2t}DE}{\frac{1}{t}DE}=\frac{\sqrt{2}}{2}$，
$\therefore$点$G$的运动速度是点$E$运动速度的$\frac{\sqrt{2}}{2}$倍，
故④错误，
故选：$C$.