For one known (invertible) function that does: $$f:H^{\otimes 2n}\times H^{\otimes 2n}:\ket{x}\ket{0}\mapsto\ket{x}\ket{y}$$$$f:H^{\otimes 2n}\times H^{\otimes 2n}:|x⟩|0⟩\mapsto|x⟩|y⟩$$ I want to find a similar (invertible) function that does: $$g:H^{\otimes 2n}\times H^{\otimes 2n}:\ket{0}\ket{y}\mapsto\ket{x}\ket{0}$$$$g:H^{\otimes 2n}\times H^{\otimes2n}:|0⟩|y⟩\mapsto|x⟩|0⟩$$ so that $f \ast g$ maps $\ket{0}\ket{y}$$|0⟩|y⟩$ to $\ket{x}\ket{y}$$|x⟩|y⟩$.
To do so, I prepare a state $\ket{0, y_x}$$|0, y_x⟩$ and do the following operations: $$\ket{0}\ket{y_x} \mapsto \frac{1}{\sqrt{2^{n}}}(\sum^{2^n-1}_{k\,=\,0}\ket{k})\ket{y_x} = \frac{1}{\sqrt{2^{n}}}\sum^{2^n-1}_{k\,=\,0}\ket{k}\ket{y_x} $$$$|0, y_x⟩ \mapsto \frac{1}{\sqrt{2^{n}}}(\sum^{2^n-1}_{k\,=\,0}|k⟩)|y_x⟩ = \frac{1}{\sqrt{2^{n}}}\sum^{2^n-1}_{k\,=\,0}|k⟩|y_x⟩ $$ Now apply $f^{-1}$ to $\ket{k}\ket{y_x}$$|k⟩|y_x⟩$, which gives: $$\frac{1}{\sqrt{2^{n}}}\sum^{2^n-1}_{k\,=\,0}f^{-1}(\ket{k}\ket{y_x})=...+\frac{1}{\sqrt{2^{n}}}\ket{x}\ket{0}$$$$\frac{1}{\sqrt{2^{n}}}\sum^{2^n-1}_{k\,=\,0}f^{-1}(|k⟩|y_x⟩)=...+\frac{1}{\sqrt{2^{n}}}|x⟩|0⟩$$ I verfied that only when $k=x$, the second register of $f^{-1}(\ket{k}\ket{y_x})$$f^{-1}(|k⟩|y_x⟩)$ contains $\ket{0}$$|0⟩$. Denote $\ket{\varphi}$$|\varphi⟩$ as the state of second register. The only information I know, is that $\ket{\varphi}$$|\varphi⟩$ contains $\ket{0}$$|0⟩$ with amplitude $\frac{1}{\sqrt{2^{n}}}$. Is that possible that I can use amplitude amplification to significantly increase the amplitude of $\ket{0}$$|0⟩$?