 Solution

Choose $z$ normal to the sheet and $x$ along the in-plane projection of the rotation axis. A uniform square has $I_x=I_y=m{a}^2/12$ and $I_z=m{a}^2/6$. Since ${\omega }_x={\omega }_z=\omega /\sqrt{2}$, the two nonzero angular-momentum components are $J_x=I_x\omega /\sqrt{2}$ and $J_z=I_z\omega /\sqrt{2}$. ThereforeAngular momentum is not parallel to the prescribed rotation axis. Multiplying the moment of inertia about that axis by $\omega$ would give the projection of angular momentum on the axis, not its magnitude.