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

Hint: $||x||^2_2 = x^H x$. What is $(Ax)^H$?
