Let $X$ be a Noetherian regular scheme, and $\mathcal{E}$ a locally free sheaf of rank $n+1\geq 2$ on $X$. Let $\pi:\mathbb{P}(\mathcal{E})\to X$ be the natural morphism. Then for every $x\in X$, the fiber of $\pi$ over $x$ is isomorphic to $\mathbb{P}_{k(x)}^n$. So if $\mathcal{L}$ is an invertible sheaf on $\mathbb{P}(\mathcal{E})$, then for every $x$ we get a unique integer $n_x=n_x(\mathcal{L})$ such that the restriction of $\mathcal{L}$ to $\mathbb{P}_{k(x)}^n$ is $\mathcal{O}_{\mathbb{P}_{k(x)}^n}(n_x)$. Then is it true that $x\mapsto n_x$ is constant? How could one prove this elementarily?
Some thoughts: if $U=\operatorname{Spec}A$ is an affine open subset which trivializes $\mathcal{E}$, then $\pi^{-1}(U)\cong \mathbb{P}_A^n$. So if we can show the statement in this much simpler case, then we already obtain that $x\mapsto n_x$ is locally constant.