I know that : If $y$ is a principal axis $\iff$ you can express the angular momentum with respect to a point $Y$ on that axis (if the solid is rotating around that axis of course) with the formula : $\vec{L_Y} = I_y\vec{\omega}$, where $I_y$ is the moment of inertia with respect to the axis $y$ (else, you would have to use a tensor of inertia)(at least, that's what I learnt at school).
Now consider that $z_G$ is a principal axis that goes through the center of mass $G$ of a solid (of mass $m$) and let $I_{z_G}$ be the moment of inertia with respect to that axis. The parallel axis theorem gives us the moment of inertia with respect to any axes $z$ parallel to $z_G$ : $I_z = I_{z_G} + md^2$, where $d$ is the perpendicular distance between the two axes. So if the solid is rotating around one of those axes $z$ with an angular velocity $\omega$, the angular momentum with respect to a point $Z$ on $z$ would be $\vec{L_Z} = I_z\vec{\omega}$, which would mean $z$ is a principal axis of inertia. Am I right ?
Thanks.