Unweighted triangle-orbit counts miss the derived homotopy factor

Let $S=k$ concentrated in degree $0$, let $x=S$, $y=S[1]$, and $z=S\oplus S[1]$. Since $\mathrm{Hom}(S,S[1])=0$ and $\mathrm{Hom}(S,S)=k$, every map $f:S\to z$ is $f_a=(a,0)$ for a unique $a\in k$. Its cone is isomorphic to $S[1]$ exactly when $a\mathrm{\ne }0$: for $a\mathrm{\ne }0$ the first component is an isomorphism and cancels in the cone, while for $a=0$ the cone has nonzero degree-zero cohomology. Hence $\mathrm{\mid }[x,z{]}_y\mathrm{\mid }=q-1$ and $\mathrm{\mid }\mathrm{Aut}(x)\mathrm{\mid }=q-1$, so $n_{x,y}^z=1$.

In this category ${\mathrm{Ext}}^{-1}(x,z)=\mathrm{Hom}(S,z[-1])\cong k$, while ${\mathrm{Ext}}^{-i}(x,z)=0$ for $i\ge 2$ and ${\mathrm{Ext}}^{-i}(x,x)=0$ for all $i>0$. Thus the higher-Ext factor in [F1] is $q^{-1}$ and the Toën coefficient is $g_{x,y}^z=q^{-1}\mathrm{\ne }1=n_{x,y}^z$. The claimed equality is false.