Kirszbraun's theorem for $\mathbb{R}^2$ states the following:

Given any set $S\subset \mathbb{R}^2$ and any Lipschitz function $f:S\rightarrow \mathbb{R}^2$ with Lipschitz constant $k$, $0< k< \infty$, for any set $F$ which contains $S$ there exists a function $\tilde f:F\rightarrow C$ such that $Lip(\tilde{f})=k$, $\tilde{f}|_{S}=f$ and $C$ is any closed convex set containing $f(S)$.

I'm wondering if something similar could hold true for bi-Lipschitz functions, for example is the following claim true?

Given any two discrete sets of $n\ge 3$ linearly independent points $S_1,S_2\subset \mathbb{R}^2$, and any bi-Lipschitz function $f:S_1\rightarrow S_2$ with bi-Lipschitz constant $k$, $0<k<\infty$, there exists a function $\tilde{f}:C_1\rightarrow C_2$ which extends $f$ and has bi-Lipschitz constant $k$, where $C_i$ is the convex envelope of $S_i$.

This problem seems more complicated than what I expected since the only article I've found on this subject is this https://arxiv.org/pdf/1110.6124.pdf which computes the bi-Lipschitz extension only in the case $S$ is the border of the unit square and the extended function has the bi-Lipschitz constant multiplied by a "big" constant factor.

User YCor proved that my claim is false in general, but do you know any other reference on this problem? In particular **is my claim true when $S$ is composed of 3 linearly independent points?** (even this doesn't seem trivial to me)