Action of $Aut(X)$ on $Coh(X)$

Question

I was reading about Bondal and Orlov reconstruction theorem. In particular that for a smooth variety with ample or anti-ample canonical bundle $\mathrm{Aut}(D^b(X)) \cong \mathbb{Z} \times (\mathrm{Pic}(X) \rtimes \mathrm{Aut}(X))$. This made me wonder whether given an automorphism $f$ of a variety $X$, I can always find a coherent sheaf $\mathcal{F}$ such that $f_* (\mathcal{F}) \ncong \mathcal{F}$. Is it true? Intuitively it seems true to me, but how to show it? For sure I have to look to something bigger than line bundles, since on projective space line bundles are classified, up to isomorphism, by degree.

Edit: for seeing that no nontrivial automorphism acts nontrivially we can look at skyscraper sheaves of points. Is there we can say in general about vector bundles?

Answer 1

If $f$ is the identity, then of course $f_*(\mathcal F)\cong \mathcal F$. So you should assume that $f \neq \mathrm{id_X}$. But then one has that $f(P) \neq P$ for some $P \in X$, in which case you can take $\mathcal F$ to be the skyscraper at $P$.