Line bundles pullback trick I'm trying to prove a statement about line bundles, and the following question is crucial to complete a possible proof that I have in mind. Could you give me any hint please?
Let $A$ and $B$ two schemes over the same base and $A\times B$ their fiber product, with canonical projection $\pi:A\times B \to A$.
Consider two line bundles $F$ and $G$ over $A\times B$ and suppose we have a morphism $f:A\to A$ such that
$$ (\;f\times id_B)^*F = G. $$
Is it possible, then, to find a line bundle $H\;$ and/or $\;H'$ on $A$ such that
$$ F\cong G\otimes\pi^*H \qquad \text{and/or}\qquad G\cong F\otimes\pi^*H' \quad ? $$
My thoughs:
Intuitively, since $f\times id_B$ acts trivially only on $B$, it should be possible to turn $F$ into $G$ just by acting on $A$. But can we do it by means of a functor $\square\otimes\pi^* H$ for some line bundle $H$ ?

I think there is an easy counter-example. Let $A=B$ be an elliptic curve and $F=\mathcal O(\Delta )$ where $\Delta$ is the diagonal. Let $f=-1_A$ be the involution induced by taking inverses on $A$. If $G^{-1}\otimes F= (f\times id_B)^*F^{-1}\otimes F\cong \pi ^*H$, then restricting to $A\times 0_B$ we get $H=\mathcal O _A$, but restricting to $A\times b$ for a general point $b\in B=A$ we have $\mathcal O _A(b-(-1_A)^*b)\not \cong \mathcal O _A$.   
