A linear transformation that maps $|0\rangle$ to $ \alpha |0\rangle+ \beta |1\rangle$ and $|1\rangle$ to $\gamma|0\rangle +\delta |1\rangle$ is unitary if $|\alpha|^2+|\beta|^2=1$ and $|\gamma|^2+|\delta|^2=1$ and $\alpha \gamma*+\beta \delta^*=0$. Here $a^*$ denotes the complex conjugate of $a$.
Show that the map that maps $|0\rangle$ to $ \alpha^* |0\rangle+ \gamma^* |1\rangle$ and $|1\rangle$ to $\beta^*|0\rangle +\delta^* |1\rangle$ is unitary and the inverse of the previous.
I was able to show that the latter is the inverse of the previous but I was not able to show that the later is unitary.
Asserting that that transformation is unitary is equivalent to asserting that$$\begin{bmatrix}\alpha&\beta\\\gamma&\delta\end{bmatrix}.\begin{bmatrix}\overline\alpha&\overline\gamma\\\overline\beta&\overline\delta\end{bmatrix}=\operatorname{Id}.$$But then each of those matrices is the inverse of the other one and therefore$$\begin{bmatrix}\overline\alpha&\overline\gamma\\\overline\beta&\overline\delta\end{bmatrix}.\begin{bmatrix}\alpha&\beta\\\gamma&\delta\end{bmatrix}=\operatorname{Id},$$which is equivalent to what you want to prove.