$\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Proj{Proj}\DeclareMathOperator\Pic{Pic}$I have a question about an example for a line bundle not admitting a $G$-linearization from Mumford's GIT, page 33:

We consider the action of $\PGL(n+1)$ on projecive space $\mathbb{P}^n= \Proj k[X_0,\dotsc, X_n]$. Observe that $\PGL(n+1)$ is given as the open affine subscheme of

$$ \mathbb{P}^{n^2+2n} = \Proj k[a_{00},\dotsc, a_{0n}; a_{10}, \dotsc , a_{nn}]$$

complementary to the determinants hypersurface $\det(a_{ij})=0$. The action morphism $ \sigma: \PGL(n+1) \times \mathbb{P}^n \to \mathbb{P}^n $ is determined by

\begin{gather*} \sigma^*(\mathcal{O}_{\mathbb{P}^n}(1)) \cong p_1^*(\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)) \otimes p_2^*(\mathcal{O}_{\mathbb{P}^{n}}(1)) \\ \sigma^*(X_i)= \sum_{j=0}^n p_1^*(a_{ij}) \otimes p_2^*(X_j) \end{gather*}

where $p_1, p_2$ are projections canonical projections.

Mumford claims that $\mathcal{O}_{\mathbb{P}^n}(1) $ admits no $\PGL(n+1)$-linearization, because the restriction of $\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$ to the open subscheme $\PGL(n+1)$ has order $n+1$ in $\Pic[\PGL(n+1)]$, and is therefore not trivial.

My question is why the fact that $\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$ restricted to the affine open $\PGL(n+1)$ is not trivial, implies that $\mathcal{O}_{\mathbb{P}^n}(1) $ admits no $\PGL(n+1)$-linearization?

[Indeed, $\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$ has order $n+1$ in $V$ because $\Pic(\mathbb{P}^{n^2+2n}) \to \Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a_{ij}))= \Pic(\PGL(n+1))$ induces an isomorphism $\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a_{ij})) \cong \mathbb{Z}/(\deg(\det(a_{ij}))\mathbb{Z}$.]

To turn it another way round, why if $\mathcal{O}_{\mathbb{P}^n}(1) $ would admit a $\PGL(n+1)$-linearization, then the restriction of $\mathcal{O}_{\mathbb{P}^{n^2+2n}}(1)$ to $\PGL(n+1)$ must be trivial? I conjecture that this argument can somehow reduced to an easy comparison of orders of group elements in groups $\Pic(X)$, $\Pic^G(X)$ but I do not see how it can be directly related.

Maybe it somehow helps to know that we have always a morphism of groups $ \Pic^G(X) \to \Pic(X)$ which is not neccessarily injective.