I know that $$ A| v \rangle = \sum _n e^{i\alpha n} | n \rangle $$ where $A$ is an unitary operator, and $ \left \{ |n\rangle \right \} $ is an orthonormal complete basis.
In that case, is it true that $$ A^{\dagger}| v \rangle = \sum _n e^{-i\alpha n} | n \rangle$$ ?
If it is true, how can I prove it?
$$\langle v| A^{\dagger} = \sum _n e^{-i\alpha_n} \langle n |$$
I don't think that your version is true.
Consider $$A= \left[ \matrix { 0&i\\ 1&0 } \right]$$
$$A^{\dagger}= \left[ \matrix { 0&1\\ -i&0 } \right]$$
and apply them to the vector $\left[\matrix{1\\0}\right]$ then express in the standard basis. $$A \left[\matrix{1\\0}\right] = e^{i ( 0 )}\left[\matrix{0\\1}\right]$$ but $$A^{\dagger} \left[\matrix{1\\0}\right] = e^{i (- \frac{\pi}{2} )}\left[\matrix{0\\1}\right]$$