Consider complex smooth hypersurfaces $X\subset\mathbb{P}^n$ and $Y\subset\mathbb{P}^m$ for $m,n\geq 4$ and a morphism $f\colon X\rightarrow Y$ which satisfies one of the properties


1. $f\_\*\mathcal{O}\_X\cong\mathcal{O}\_Y$ and $f\_\*\mathcal{O}\_X(1)\cong\mathcal{O}\_Y(1)$
2. $f\_\*\mathcal{O}\_X(i)\cong\mathcal{O}\_Y(i)$ for $i=0,1,\dots,b$ for some $b>1$
3. $f\_\*\mathcal{O}\_X(i)\cong\mathcal{O}\_Y(i)$ for all $i\in\mathbb{Z}$


Here $\mathcal{O}\_X(1)$ and $\mathcal{O}\_Y(1)$ are the natural polarizations induced from $\mathbb{P}^n$ and $\mathbb{P}^m$ respectively.


Do we have (1) $\Rightarrow$ (2) $\Rightarrow$ (3)? In the case (3), can we say that $f\_\*:D^b(X)\rightarrow D^b(Y)$ induces an isomorphism of the full subcategory of line bundles?



morphisms between smooth hypersurfaces that preserve many line bundles
Your assumption about $f$ is very strong, it implies that $X \cong Y$ and $f$ is an isomorphism.


Indeed, let $Z \subset X$ be a general fiber of $f$. Then the assumption that $f\_\*\mathcal{O}\_X(1)$ is a line bundle implies that
$$
\dim H^0(Z, \mathcal{O}\_X(1)\vert\_Z) = 1.
$$
But $\mathcal{O}\_X(1)\vert\_Z$ is a very ample line bundle on $Z$, hence $Z$ is a single point. Thus, $f$ is birational. If it is not an isomorphism, the relative Picard group must be nontrivial, but since
$$
\mathrm{Pic}(X) = \mathrm{Pic}(Y) = \mathbb{Z},
$$
this is impossible, hence $f$ is an isomorphism.


