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.]

Corrected after Noam's comments below

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 $\mathrm{PO}(6,\mathbb{C}) = \mathrm{O}(6,\mathbb{C})/\{\pm I_6\}$, which has two components. Its identity component, by the usual exceptional isomorphism (i.e., $A_3=D_3$), is $$\mathrm{PSO}(6,\mathbb{C}) = \mathrm{SO}(6,\mathbb{C})/\{\pm I_6\}=\mathrm{PSL}(4,\mathbb{C}).$$

[See Noam's comments below, which point out that my original answer of $\mathrm{PSO}(6,\mathbb{C})$ was not correct and that my first attempted fix also had an error. I'm leaving them there because they might help some else avoid making my mistake in the future. ]

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 $\mathrm{PSL}(4,\mathbb{C})$.

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.]