Intuitive argument for the transitivity of $PGL(n+1)$ acting on $\mathbb{P}^n$

Question

In spherical geometry we can consider the action of $\operatorname{O}(n+1)$ on the unit sphere $\mathbb{S}^n$. It's easy to see that this action is transitive, because for any two points $x,y \in \mathbb{S}^n$ we can take $x$ to $y$ by a rotation of the sphere (there's a great circle through any two points on the sphere and a translation along this circle corresponds to a rotation around the pole axis). In the case of projective geometry the set is $X = \operatorname{P}^n(\mathbb{R}) = \{\text{1-dim subspaces of } \mathbb{R}^{n+1}\}$ and the group acting on this set is $G = \operatorname{PGL}(n+1, \mathbb{R}) = \operatorname{GL}(n+1, \mathbb{R})/\mathbb{R}^{*}$ where $\mathbb{R}^{*} = \mathbb{R}\setminus\{0\}$. This action is transitive too, but is there also an intuitive explanation for this fact?