$A^H=A^{-1}$ implies $\|x\|_2 = \|Ax\|_2$ for any $x\in \mathbb{C}^n$ Show that the following conditions are equivalent.
1) $A\in \mathbb{C}^{n\times n}$ is unitary. ($A^H=A^{-1}$)
2)  for all $x \in \mathbb{C}^n$, $\|x\|_2 = \|Ax\|_2$, where $\|x\|_2$ is the usual Euclidean norm of $x \in \mathbb{C}^n. $
I am totally lost in this problem, I appreciate any hint. And here is my argument.
From $1\to 2$, we get $A^{H}A=I$. By this problem "Prove that $\|A\|_2 = \sqrt{\|A^* A \|_2}$", I can say $\|Ax\|_2=\sqrt{\|A^HAx\|_2}$. Therefore, I get $\|Ax\|_2^2=\|x\|_2$.
I also have problem to show the equality in this problem "Prove that $\|A\|_2 = \sqrt{\|A^* A \|_2}$". 
 A: For the equality
$$\|Ax\|_2 = \left((Ax)^TAx\right)^{1/2} =  \left(x^TA^TAx \right)^{1/2} = (x^TIx)^{1/2} = (x^Tx)^{1/2} = \|x\|. $$ 
A: Hint: $||x||^2_2 = x^H x$. What is $(Ax)^H$?
A: 1 $\implies$ 2: since $A^HA=I$, $$\|Ax\|_2^2 = (Ax)^H(Ax) = x^H A^H Ax = x^H x = \|x\|_2^2.$$

2 $\implies$ 1:
$A^H A=I$ is equivalent to showing $\|A_j\|_2^2=1$ and $A_i^H A_j=0$ for each $i,j \in \{1,\ldots,n\}$, where $A_j$ is the $j$th column of $A$.
Applying statement 2 to $x=e_j$ for each standard basis vector $e_j$, we get $\|A_j\|_2=\|e_j\|_2=1$.
Applying statement 2 to $x=e_i+e_j$ for $i \ne j$ implies $$2 + 2A_i^H A_j = \|A_i\|_2^2 + 2 A_i^H A_j + \|A_j\|_2^2 = \|A_i+A_j\|_2^2 = \|e_i+e_j\|_2^2=2,$$
so $A_i^H A_j=0$.
A: Notice that for any complex $n\times n$ matrix $A$,
$$ \| Ax \|^2 = \langle Ax, Ax \rangle = \langle x, A^H A x \rangle, \qquad \forall x \in \Bbb{C}^n. $$
So the implication $1) \Rightarrow 2)$ is straightforward. For the other direction, the following lemma is useful:

Lemma. If $A$ is a $n\times n$ matrices such that $\langle x, Ax\rangle = 0$ for all $x \in \Bbb{C}^n$, then $A = 0$.

Then the implication $2) \Rightarrow 1)$ immediately follows by applying the lemma to the matrix $A^H A - I$.

Proof of Lemma. Plug $x = \alpha \mathrm{e}_k + \beta \mathrm{e}_l$, where $\alpha, \beta \in \Bbb{C}$ and $l \neq k$. Then
$$0 = |\alpha|^2 A_{kk} + \bar{\alpha}\beta A_{kl} + \alpha\bar{\beta}A_{lk} + |\beta|^2 A_{ll}. $$
Plugging $(\alpha, \beta) = (1, 0)$, we find that all the diagonal entries of $A$ are zero. Plugging $(\alpha, \beta) = (1, 1)$ and $(\alpha, \beta) = (1, i)$ respectively, we obtain a system of equations
$$ A_{kl} + A_{lk} = 0, \qquad A_{kl} - A_{lk} = 0. $$
Solving this equations shows that all the off-diagonal entries of $A$ are zero. ////
Remark. The lemma above is no longer true on real vector spaces. Consider $\frac{\pi}{2}$-rotation, for instance.
