Let $F$ be a field. A collineation of $F^3$, considered as the projective plane $FP^2$, is a permutation of lines and planes of $F^3$ that pass through the origin that preserves incidence. It's straightforward to check that any bijective linear operator on $F^3$ induces a collineation.
A collineation is said to have center $U$, where $U$ is a line, if it fixes $U$ and all planes containing it. Dually, it's said to have axis $V$, where $V$ is a plane, if it fixes $V$ and all lines in $V$.
We say that $FP^2$ is $(U, V)$-transitive if for any lines $U_1, U_2$ that are not in $V$, that are distinct from $U$, and that are coplanar with $U$, there is a collineation that has center $U$, axis $V$, and maps $U_1\mapsto U_2$. I want to show that $FP^2$ is $(U, V)$ transitive for any line $U$ and plane $V$.
First, I assume that $U\leq V$. Then, there's a bijective linear operator that acts as the identity on $V$ and sends $U_1\mapsto U_2$. This obviously has axis $V$, but I'm not sure how to prove nicely that this linear operator also fixes all planes through $U$. Is there a quick proof of this fact?
Second, there's the case where $U$ is not a subspace of $V$. In this case, I'm not sure how to build a bijective linear operator as required. There are actually many choices that fix $U$ and send $U_1\mapsto U_2$ (since we can scale along lines), and I still need to pick a third basis vector and its image so that $V$ and all its lines are fixed. Is there a nice way to do this? Note that $F$ is an arbitrary field, so I'm not sure how much I have to work with.