Autoequivalences of $\operatorname{Coh}(X)$

Question

Let $X$ be a smooth projective variety over an algebraically closed field $k$ of characteristic zero.

Is there a description of $\operatorname{Aut}(\operatorname{Coh}(X))$, i.e. the autoequivalences of the category $\operatorname{Coh}(X)$?

Clearly it contains $\operatorname{Aut}(X)\ltimes\operatorname{Pic}(X)$ as a subgroup.

Answer 1

In fact, the story with autoequivalences of $\mathrm{Coh}(X)$ is much more simple than the story of $\mathrm{D}^b(X)$. Indeed, the structure sheaves of points of $X$ are intrinsically determined as simple objects of $\mathrm{Coh}(X)$ (i.e., objects that have no nontrivial subobjects). Therefore, any autoequivalence must take the structure sheaf of a point to the structure sheaf of another point. It follows that the bimodule over $\mathrm{Coh}(X)$ that gives the autoequivalence is a line bundle twist of the structure sheaf of a graph of an automorphism of $X$, hence the autoequivalence is a composition of an automorphism and a twist.

Answer 2

A partial answer: according to a theorem of Bondal and Orlov, if the canonical bundle $\omega$ or the anticanonical bundle $-\omega$ of your variety $X$ is ample, then the bounded derived category $D^b(X)$ of coherent sheaves has $\operatorname{Aut}(X)\ltimes(\operatorname{Pic}(X)\oplus \mathbb{Z})$ as its group of exact autoequivalences. Here the $ \mathbb{Z}$ term comes from the operation of shifting the complex, so certainly nothing more than $\operatorname{Aut}(X)\ltimes\operatorname{Pic}(X)$ can come from autoequivalences of $Coh(X)$ itself.