Is there a transformation from $n-$sphere to a $(n+1)-$cube?

Question

Does there exist a smooth transformation (probably isometry) from $n-$sphere $\Bbb S^n$ to a $(n+1)-$cube (with rounded corners)? Note that $2$-cube is a square on $\Bbb R^2$ and $3$-cube is a usual cube (not filled) on $\Bbb R^3$ and so on.

Answer 1

Let $X$ denote the $n$-dimensional cube, embedded in $\mathbb{R}^{n+1}$ such that the origin is in the interior of the cube. The map $$\varphi:X\to S^n,\quad x\mapsto\frac{x}{\|x\|}$$ is a diffeomorphism. There is no isometry, as the cube and the sphere don't have the same curvature.