$A^H=A^{-1}$ 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 $\|A\|_2 = \sqrt{\|A^* A \|_2}$". 

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 $A_i^H A_j=0$.
