 Hadamard conjugation of a controlled gate

For a controlled unitary gate $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$, Hadamard gates on the target before and after the gate produce $C(HVH)$. Proof: the control-zero branch applies $HIH=I$, while the control-one branch applies $HVH$. Linearity extends the identity to arbitrary quantum states. Since $C(H)C(V)C(H)=C(HVH)$ too, a pair of controlled Hadamard gates surrounding the same target operation can be replaced by ordinary target Hadamard gates. No single controlled Hadamard gate is thereby identified with an ordinary one.