Suppose that we are given a smooth projective variety $X$ with a *full exceptional collection* of vector bundles $(F\_1, F\_2, \ldots, F\_k)$ in $D^b(X)$ and two vector bundles $E\_1$, $E\_2$ on $X$. Consider the following statement:
$$ \text{If }H^i(X, E\_1\otimes F\_j) = H^i(X, E\_2\otimes F\_j)\text{ for all }i, j\text{, then } E\_1=E\_2. $$


**Question 1.** Is the above true?


If no,


**Question 2.** What if we assume that the collection is strong?


**Question 3.** What if we consider all twists by the ample sheaf $\mathcal{O}(1)$, i. e.
$H^i(X, E\_1\otimes F\_j(t)) = H^i(X, E\_2\otimes F\_j(t))$ for $i, j$ and all $t$?


**Question 4.** Are there any known cohomological criteria for isomorphism?


If yes,


**Question 5.** Does it suffice to assume that $(F\_1, F\_2, \ldots, F\_k)$ generate the derived category, *without* them forming an exceptional collection?


**Remark.** Take $X=\mathbb{P}^n$ and $F\_i = \mathcal{O}(i)$ or $F\_i = \Omega^i(i)$. Then by the Beilinson spectral sequence, the cohomology groups $H^i(X, E\otimes F\_i)$ 
*would* determine $E$ unambiguously *if* we knew the differentials. 



Exceptional collections and cohomological criteria for isomorphism
It is hard to think of a general way to obtain an isomorphism of vector bundles from several isomorphisms of cohomology spaces. In particular, there can be moduli of vector bundles, I presume, even over projective spaces. But an isomorphism of cohomology only means an equality of dimensions, and dimensions are discrete invariants, taking values in the integers.


It would be an entirely different matter if the question were, is a given morphism of vector bundles an isomorphism provided that it induces isomorphisms of certain cohomology spaces. For this version of your question to have a positive answer, it suffices to require that (Fi) generate the derived category (due to Serre duality).


