Picard group of $G(k, n)$ saying about automorphisms

Picard group of $G(k, n)$ saying about automorphisms

Let $G(k, n)$ be the Grassmannian of $k$-dimensional subspaces of $k^{n}$,
$k$ a field, embedded in $\mathbb{P}^{N}$ by the Plücker embedding. In
Harris' Algebraic Geometry, A First Course, Theorem 10.19 states that
$$\mathrm{Aut}(G(k, n)) = \mathrm{Aut}(G(k, n), \mathbb{P}^{N}),$$ where
$\mathrm{Aut}(G(k, n), \mathbb{P}^{N}) := \{ T \in
\mathrm{Aut}(\mathbb{P}^{N}) \mid T(G(k, n)) = G(k, n) \}$. In its proof,
Harris says that it comes down to the assertion that every codimension $1$
subvariety of $G(k, n)$ is the intersection of $G(k, n)$ with a
hypersurface in $\mathbb{P}^{N}$. In other words, $\mathrm{Pic}(G(k, n))
\cong \mathrm{Pic}(\mathbb{P}^{N})$, right? How such information can say
about the automorphism group of these varieties?
Thank you!