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 do 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.

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.