$\because$多边形$ABCDEFGHIJ$为正十边形，
$\therefore$正十边形的内角为：$\frac{1}{10}\times \left(10-2\right)\times 180^{\circ}=144^{\circ}$，
$\therefore \angle AJI=\angle IAB=\angle ABC=144^{\circ}$，$AJ=JI=AB=BC$，
$\therefore \angle JAI=\frac{1}{2}\times \left(180^{\circ}-\angle AJI\right)=\frac{1}{2}\times \left(180^{\circ}-144^{\circ}\right)=18^{\circ}$，
$\therefore \angle IAB=\angle JAB-\angle JAI=144^{\circ}-18^{\circ}=126^{\circ}$，
根据正十边形的性质得：$AB$∥$JC$，
$\therefore \angle IAB+\alpha =180^{\circ}$，
$\therefore \alpha =180^{\circ}-\angle IAB=180^{\circ}-126^{\circ}=54^{\circ}$.
故答案为：$54$.

$\because$多边形$ABCDEFGHIJ$为正十边形，
$\therefore$正十边形的内角为：$\frac{1}{10}\times \left(10-2\right)\times 180^{\circ}=144^{\circ}$，
$\therefore \angle AJI=\angle IAB=\angle ABC=144^{\circ}$，$AJ=JI=AB=BC$，
$\therefore \angle JAI=\frac{1}{2}\times \left(180^{\circ}-\angle AJI\right)=\frac{1}{2}\times \left(180^{\circ}-144^{\circ}\right)=18^{\circ}$，
$\therefore \angle IAB=\angle JAB-\angle JAI=144^{\circ}-18^{\circ}=126^{\circ}$，
根据正十边形的性质得：$AB$∥$JC$，
$\therefore \angle IAB+\alpha =180^{\circ}$，
$\therefore \alpha =180^{\circ}-\angle IAB=180^{\circ}-126^{\circ}=54^{\circ}$.
故答案为：$54$.