EDIT: This answer mistakenly assumes that $A$ is a square $n \times n$ matrix.
I assume we're working over $\mathbb R$. We claim that $A$ must be a scalar multiple of an orthogonal matrix.
First, we prove that if $\| x \| = \| y \|$, then $\| A x \| = \| A y \|$. Towards a contradiction, assume that $\| x \| = \|y \|$ and $\| A x \| \neq \| A y \|$. Without loss of generality, we can assume that $\| Ax \| < \|A y \|$. Moreover, it is clear that both $x$ and $y$ are nonzero. (Why?) Now, fix a number $\beta$ such that $ 1 < \beta < \frac{\| A y \|}{\| Ax \|}. $ Then, defining $z = \beta x$, it is clear that
This is a contradiction to the hypothesis (since $\| A z \| < \| A y \|$ but \| z \| > \| y \|). Hence, if $\| x \| = \| y \|$, then $\| A x \| = \| A y \|$.
It now remains to show that $A$ is a multiple of an orthogonal matrix. Fix a unit vector $u$. Then since $\| x \| = \| (\| x \| u) \|$, it follows from (2.) that $\| A x \| = \| A (\| x \| u) \| = \| x \| \cdot \| A u \|$. Now, if $\| A u \| = 0$, then $A$ must be the zero matrix (why?) and we are already done. On the other hand, assuming $\| A u \| > 0$, it is easy to see that the matrix $ B := \frac{1}{\| A u \|} A $ is a linear isometry and hence orthogonal.