Identity and zero transitions obstruct a diagram comparison

If $F:A\Rightarrow B$ is a natural isomorphism, its components are invertible by [F1]. Naturality on $u$ says $z{F}_0=F_1$, hence $z=F_1{F}_0^{-1}$ is invertible. Conversely, if $z$ is invertible, choose $F_0={\mathrm{id}}_c$ and $F_1=z$. The only nonidentity naturality equation is then satisfied, and the inverse components ${\mathrm{id}}_c,z^{-1}$ satisfy the reverse equation. Thus these components give inverse natural transformations.

For a small dg category $\mathcal{C}$, let $S(\mathcal{C})$ be the set of isomorphism classes of objects of $H^0(\mathcal{C})$. A dg functor induces a function on these sets, preserving identities and composition, so $S$ is a functor to sets in the ambient larger size convention. A quasi-equivalence induces a bijection: essential surjectivity gives surjectivity, and an isomorphism between two image objects lifts by fullness on $H^0$ to inverse arrows, whose inverse equations follow by faithfulness. Therefore [F2] factors $S$ through $\mathcal{D}$. In particular an invertible arrow of $\mathcal{D}$ must induce a bijection on $S$.

The complex $\mathbb{C}$ concentrated in degree zero and the zero complex have distinct classes in $S(c)$: its degree-zero endomorphism cohomology is $\mathbb{C}$, with nonzero identity, whereas every map to or from the zero complex is zero. The map $S(z)$ sends both classes to the zero-complex class, so it is not injective. Step 1.2 shows that $z$ is not invertible in $\mathcal{D}$. The two diagrams are genuine functors since the only compositions to check involve identities; their fibres agree literally. Step 1.1 now rules out a natural isomorphism, as claimed.