In the context of projective complex geometry (say), let $E$ be an elliptic curve. Write $\mathcal{O}_{\Delta}$ for the diagonal on $E \times E$ and let $\varphi$ be a non-trivial member of $\mathrm{Ext}(\mathcal{O}_{\Delta}, \mathcal{O}_{\Delta})$.
This gives a non-trivial morphism $\varphi : \mathcal{O}_{\Delta} \to \mathcal{O}_{\Delta}[2]$, which induces a natural transformation $\Phi_{\varphi} : \Phi_{\mathcal{O}_{\Delta}} \to \Phi_{\mathcal{O}_{\Delta}[2]}$ (where $\Phi_{\ast}$ denotes the FM transform with kernel "$\ast$"), in this case, $\Phi_{\varphi} : \mathrm{id} \to [2]$. On complexes, this is a map $\mathcal{F}^{\ast} \to \mathcal{F}^{\ast}[2]$.
I'm trying to prove that such a map is always trivial, as indicated in the discussion following Orlov's theorem in Huybrechts - FM transforms in geometry, chapter 5. This will then give a contradiction.
Of course, for a sheaf concentrated somewhere, $\mathrm{Hom}(\mathcal{F}, \mathcal{F}[2]) = \mathrm{Ext}^2(\mathcal{F}, \mathcal{F})$ is zero (since $E$ is of dimension $1$). On complexes however, the best you can do seems to be to apply the two $\mathrm{Ext}$ spectral sequences and reduce to $\mathrm{Hom}(\mathcal{F}^q, \mathcal{F}^{q+2}) = \mathrm{Ext}^1(\mathcal{F}^{q+1}, \mathcal{F}^{q+2}) = 0$ for all $q$.
As far as I can tell, this may be false: take for instance the complex
$\cdots \to 0 \to \mathcal{O} \to 0 \to \mathcal{O} \to 0 \to \cdots$
and the morphism to its shift by two given by some non-trivial map $\mathcal{O} \to \mathcal{O}$.
I would be thankful if you would give me some help with this question.
If $E$ is a curve, any object of the derived category $D(E)$ is quasiisomorphic (noncanonically) to the direct sum of its cohomology sheaves: $$ F \cong \bigoplus_{i = -\infty}^{\infty} H^i(F)[-i]. $$ Any morphism of functors commutes with direct sums and shifts, so the statement you up to reduces to the statement about sheaves.