Contracting a set to a ball

Question

$\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$

Question 1: Let $S$ be a nonempty measurable subset of $\R^n$. Let $B$ be a closed ball in $\R^n$ such that $m(B)=m(S)$, where $m$ is the Lebesgue measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $B$?

Question 2: If such a map exists, can we make it measure-preserving?

Question 1a: Let $S$ be a nonempty measurable subset of $\S^{n-1}$, the unit sphere in $\R^n$. Let $C\subseteq\S^{n-1}$ be a closed spherical cap such that $m(C)=m(S)$, where $m$ is the Haar measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $C$? (The metric on $\S^{n-1}$ with respect to which the $1$-Lipschitz condition is considered is either the geodesic metric on $\S^{n-1}$ or, equivalently, the one induced by the Euclidean metric on $\R^n$.)

Question 2a: If a map described in Question 1a exists, can we make it measure-preserving?

A complete and correct answer to any one of these four questions will be considered a complete and correct answer to the entire post.

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Related, but different, questions and answers can be found on this MathOverflow page of 2018.

Answer 1

Trivial "No" to all: Take the union $S$ of two disjoint balls $B_1$ and $B_2$ of diameters $d$ and $\sqrt{1-d^2}$ respectively. If $f$ maps $S$ to the ball $B$ of diameter $1$, then $f(B_1)$ has diameter $\le d$. If $d$ is small enough, then $B\setminus f(B_1)$ still contains two opposite points on the circumference, so the diameter of it is still $1$ and $f(B_2)$ has no chance to get anywhere close to covering it.