The subgroup of $\mathrm{PSL}(6,\mathbb{C})$ acting in its usual way on $\mathbb{CP}^5$ that preserves the quadric hypersurface that is $G(2,4)\subset \mathbb{CP}^5$ is $\mathrm{PSO}(6,\mathbb{C}) = \mathrm{SO}(6,\mathbb{C})/\{\pm I_6\}$, which, by the usual exceptional isomorphism (i.e., $A_3=D_3$), is the same as $\mathrm{PSL}(4,\mathbb{C})$.  

[However, see Noam's comment below (which points out that I didn't quite consider the right group) and my response.]