B

如图，连接$AC$、$BD$交于点$E$，过点$A$作$MN\bot y$轴于点$M$，过点$C$作$CN\bot y$轴于点$N$，

$\therefore \angle DNC=\angle AMD=90^{\circ}$，
$\therefore \angle ADM+\angle DAM=90^{\circ}$，
$\because$四边形$ABCD$是正方形，$CD=AD$，$\angle ADC=90^{\circ}$，
$\therefore \angle ADM+\angle NDC=90^{\circ}$，
$\therefore \angle CDN=\angle DAM$.
在$\triangle CND$和$\triangle DMA$中
$\left\{\begin{array}{l}{∠CND=∠AMD=90°}\\{∠CDN=∠DAM}\\{CD=DA}\end{array}\right.$，
$\therefore \triangle CND$≌$\triangle DMA\left(AAS\right).$
$\therefore AM=DN=m$，$DM=CN=n$.
$\because$点$A$、$C$的横坐标分别为$m$、$n$，
$\therefore A(m$，$-m^{2}+6)$，$C(n$，$-n^{2}+6)$.
$\therefore ON=-n^{2}+6$，$OM=-m^{2}+6$，
$\because ON=OM+MD+DN$，
$\therefore -m^{2}+6+m+n=-n^{2}+6$，
$\therefore \left(m+n\right)\left(m-n\right)=m+n$.
$\because m \gt n \gt 0$，
$\therefore m+n\neq 0$.
$\therefore m-n=1$.
故选：$B$.