Proposition: An exact effect does not certify a qubit

Let

The effect $F=|0⟩⟨0|$ is preserved exactly, while

Proof. ${{\Delta }_Z}^{†}(F)=F$. The upper bound follows from ${\Delta }_Z-{\mathrm{id}}_2=({\mathrm{Ad}}_Z-{\mathrm{id}}_2)/2$ and the fact that two channels have half-diamond distance at most one. On $|+⟩⟨+|$, the output is $I_2/2$, at trace distance $1/2$ from $|+⟩⟨+|$.