如图，过点$C$作$CD\bot y$轴于点$D$，

$\therefore \angle CDB=\angle BOA=90^{\circ}$，
$\therefore \angle CBD+\angle BCD=90^{\circ}$.
$\because \triangle ABC$为等腰直角三角形，
$\therefore AB=BC$.
$\because \angle ABC=90^{\circ}$，
$\therefore \angle CBD+\angle ABO=90^{\circ}$，
$\therefore \angle BCD=\angle ABO$，
$\therefore \triangle AOB$≌$\triangle BDC\left(AAS\right)$，
$\therefore CD=OB$，$BD=OA$.
$\because A\left(2,0\right)$，$B\left(0,t\right)$，
$\therefore OA=2$，$OB=t$，
$\therefore CD=t$，$BD=2$，
$\therefore OD=OB+BD=t+2$，
$\therefore$点$C$的坐标为$\left(t,t+2\right)$，
$\therefore$点$C$关于$y$轴的对称点${C\’}$的坐标为$\left(-t,t+2\right)$.
故答案为：$\left(-t,t+2\right)$.

如图，过点$C$作$CD\bot y$轴于点$D$，

$\therefore \angle CDB=\angle BOA=90^{\circ}$，
$\therefore \angle CBD+\angle BCD=90^{\circ}$.
$\because \triangle ABC$为等腰直角三角形，
$\therefore AB=BC$.
$\because \angle ABC=90^{\circ}$，
$\therefore \angle CBD+\angle ABO=90^{\circ}$，
$\therefore \angle BCD=\angle ABO$，
$\therefore \triangle AOB$≌$\triangle BDC\left(AAS\right)$，
$\therefore CD=OB$，$BD=OA$.
$\because A\left(2,0\right)$，$B\left(0,t\right)$，
$\therefore OA=2$，$OB=t$，
$\therefore CD=t$，$BD=2$，
$\therefore OD=OB+BD=t+2$，
$\therefore$点$C$的坐标为$\left(t,t+2\right)$，
$\therefore$点$C$关于$y$轴的对称点${C\’}$的坐标为$\left(-t,t+2\right)$.
故答案为：$\left(-t,t+2\right)$.