Are automorphism of $\mathbb{P}^2$ 4-transitive? Given two set of four points, both of them not colinear, is there always $g\in Aut(\mathbb{P}^2)$ such that it sends one set two the other?
 A: Yes, given two sets of four non-colinear points you can send one set unto the other unless one set has three colinear points and the other hasn't, in which case the impossibility is clear.
Transitivity in the general  case (no three points colinear) is solved by consideration  of the notion of projective frame. See Théorème (3.2.1) here.
The remaining case of transitivity (when exactly three points are colinear in each set) is even easier.   
A: Georges gave a reference; let me sketch the proof to show that it is easy.
First, given three non-collinear points, you can always take them by a projective transformation to the three coordinate points $[1,0,0], \, [0,1,0], \, [0,0,1]$. This is just the fact that any two bases of a vector space are related by a linear transformation.
After doing that, since the fourth point is not collinear with any two of the others, it must have coordinates $[x,y,z]$ with all of $x, \, y, \,z$ nonzero. But now just apply the matrix $\operatorname{diag} \left(\frac{1}{x},\frac{1}{y},\frac{1}{z} \right)$.
