Can anyone help me with this question on vibrational modes of free base porphyrin (C20H14N4)?
If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.



Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.