Is there any diffeomorphism $x:\mathbb{R}^n\to\text{Im }x=B\_1\left(0\right)\subset\mathbb{R}^n$ such that 


1. $x$ is an orthogonal chart, i.e., the coordinate vector fields $X\_i=\partial x/\partial u\_i$ satisfy 
$X\_i\left(u\right)\cdot X\_j\left(u\right)=0$ for all $1\leq i\neq j\leq n$ at every point $u=\left(u\_1,\dots,u\_n\right)\in\mathbb{R}^n$;
2. the Euclidean norm of the differential $dx$ satisfies $$\left\|dx\left(u\right)\right\|^2\mathrel{\mathop:}=\sum\_{i=1}^n\left\|X\_i\left(u\right)\right\|^2=\rho^2\left(x\left(u\right)\right)$$ 
at every point $u\in\mathbb{R}^n$, where $\rho:B\_1\left(0\right)\to\mathbb{R}$ is given by $\rho\left(x\right)=\frac{1-\left\|x\right\|^2}{2}$. In particular, $dx$ vanishes at infinity.


The above assumptions lead to the following second-order PDE
$$\Delta x-\rho\left(x\right)x=\sum\_{i=1}^n\frac{\partial}{\partial u\_i}\ln\left(\left\|X\_i\right\|^2\right)X\_i,$$
which might be useful somehow. 


$\hspace{4pt}$ I believe that such a diffeomorphism does not exist. Actually, I can't even think of such an $x$ satisfying just condition 1. By Liouville's theorem we know that $x$ cannot be conformal. Moreover, radial diffeomorphisms typically used to show that $\mathbb{R}^n$ and $B\_1\left(0\right)$ are diffeomorphic do not satisfy condition 1 nor 2.



Global orthogonal coordinates on the open unit ball
This problem is just the classical problem of finding global Tchebychev coordinates on hyperbolic $n$-space. By Hilbert's Theorem, this is impossible when $n=2$. The problem remains open in higher dimensions, despite years of work.


Here is how one can see the reformulation: Let $J$ be the Jacobian of $x$ with respect to $u$, i.e., $\mathrm{d}x = J\,\mathrm{d}u$. Then condition 1 is that $J^TJ$ be diagonal, say, $J^TJ = \mathrm{diag}(\lambda\_1^2,\ldots,\lambda\_n^2)$, and condition 2 is that 
$$
\lambda\_1^2+\cdots+\lambda\_n^2 = (1-\|x\|^2)^2/4.
$$
Writing $\lambda\_i = \tfrac12(1{-}\|x\|^2) g\_i$ yields 
$$
g\_1^2+\cdots+g\_n^2=1\tag1
$$ 
and the above equations become
$$
(g\_1\,\mathrm{d}u\_1)^2 + \cdots +(g\_n\,\mathrm{d}u\_n)^2 
= \frac{4}{(1-\|x\|^2)^2}\bigl(\mathrm{d}x\_1^2+\cdots+\mathrm{d}x\_n^2\bigr)
\tag2
$$
The right hand side of (2) is the standard form of the conformal hyperbolic metric on the unit ball, so the question is asking for coordinates $u\_i$ and functions $g\_i>0$ on hyperbolic $n$-space satisfying equations (1) and (2). This is the classic problem of Tchebychev coordinates.


For example, when $n=2$, one can write $(g\_1,g\_2)=(\cos f,\sin f)$ for some function $f$ taking values in $(0,\tfrac12\pi)$, and then one is asking when
the metric
$$
ds^2 = \cos^2f\,\mathrm{d}u\_1^2 + \sin^2f\,\mathrm{d}u\_2^2
$$
has Gauss curvature $-1$. It is well-known that this holds if and only if $f$ satisfies the Sine-Gordon equation
$$
\frac{\partial^2f}{\partial u\_1^2} - \frac{\partial^2f}{\partial u\_2^2} 
= \cos f\,\sin f.\tag3
$$
Hilbert proved that there do not exist global coordinates $u\_i$ on the hyperbolic plane and a function $f(u\_1,u\_2)$ taking values in $(0,\tfrac12\pi)$ satisfying (3).


As I wrote, it is still not known whether global solutions (with $g\_i>0$) exist in higher dimensions, but, as Cartan showed, there do exist many local solutions. Essentially, they depend on $n(n{-}1)$ functions of one variable.


**Addendum:** Of course, if one only wants global orthogonal coordinates, that is not hard: If you first apply inversion with respect to a point on the boundary of the ball $B\_1(0)$ in $x$-space, you'll get an injective diffeomorphism $v:B\_1(0)\to\mathbb{R}^n$ whose image is a half-space. Say $v\bigl(B\_1(0)\bigr)$ is defined by $v\_n>0$ (after some translation and rotation in $v$-space). Then define
$$
(u\_1,\ldots,u\_{n-1},u\_n) = \bigl(v\_1,\ldots,v\_{n-1},\log(v\_n)\bigr).
$$
You'll then have $u:B\_1(0)\to\mathbb{R}^n$ being a bijective orthogonal diffeomorphism, and its inverse will be the diffeomorphism $x:\mathbb{R}^n\to B\_1(0)$ that you desire.


When $n=2$, there are more options: Choose a conformal diffeomorphism $z:B\_1(0)\to R\subset\mathbb{C}$ where $R = I\_1\times I\_2$ is a rectangle that is a product of two intervals $I\_i\subset\mathbb{R}$, not both equal to $\mathbb{R}$,
and let $z = v\_1 + i v\_2$ and choose any bijective diffeomorphisms $h\_i:I\_i\to\mathbb{R}$. Then $u\_1 = h\_1(v\_1)$ and $u\_2 = h\_2(v\_2)$ will provide the desired functions. Now let $x:\mathbb{R}^2\to B\_1(0)$ be the inverse of $u:B\_1(0)\to\mathbb{R}^2$.


