As explained in the comments of this answer, given a smooth Fano 3-fold of index 1 and genus $g \geq 6$, we have two semiorthogonal decompositions $$\langle \text{Ku}(X), \mathcal{E}, \mathcal{O}_X\rangle$$ and $$ \langle \text{Ku}’(X), \mathcal{O}_X, \mathcal{E}^\vee \rangle $$

obtained from effectively dualizing the rectangular Lefschetz decomposition. The functor $\mathbb{L}_{\mathcal O}(- \otimes \mathcal{O}(H) )$ in this case gives an equivalence between $\text{Ku}(X)$ and $\text{Ku}’(X)$. It is also obvious that any two exceptional collections related by a series of mutations give the same Kuznetsov component. Are there a priori any reasons in general one should expect the Kuznetsov component to be independent of the choice of exceptional collection?