One fixed point is easy. The line $AC$ gets mapped to $BD$ by the transformation $f$. Conversely $BD$ gets mapped to $CA$. So the intersection of these two lines must remain fixed.
Assume your four points form a regular square. Then the operation $f$ is a rotation by $90°$. A rotation is an Euclidean transformation, i.e. an isometry, and as such preserves the ideal circle points $[1:\pm i:0]$. For a generic quadrilateral, those points will likely have different coordinates, but it is important to take possibly complex positions into account. Any other quadrilateral is equivalent to a square under real projective transformations. Which means that the second and third fixed point will always be a pair of mutually conjugate complex points.
Now let's look at the fixed lines, which I assume is what you mean by invariant planes. One is again fairly easy. Since line $AB$ gets mapped to $BC$, that to $CD$ and that to $DA$, you can do with these lines what we did for the points. The intersection of $AB$ with $CD$ will trade places with that of $BC$ with $DA$. Which means the line joining them will be fixed.
Since projective transformations preserve incidence, the line joining two fixed points must be fixed, and dually the intersection of two fixed lines will be a fixed point. The fixed line I just described is the line joining the pair of conjugate complex fixed points. The two other fixed lines can be obtained by joining the real fixed points to one of the complex fixed points.