We say a mapping $f:\mathbb R^n\to \mathbb R^n$ be 1-Lipschitz with respect to a norm $\|\cdot\|$ if $\|f(x)-f(z)\|\le\|x-z\|$ holds for all $x,z\in\mathbb R^n$. Such a mapping are sometimes called a contraction.

I want to study the existence of non-trivial distance-preserving mappings for general norm, such that the inequality becomes an equality, i.e. $\|f(x)-f(z)\|=\|x-z\|$ for all $x,z\in \mathbb R^n$.

Obviously, there are several trivial mappings that satisfy the above condition:

• Identity mapping $f(x)=x$ and its negative counterpart $f(x)=-x$;
• Addition: $f(x)=x+b$ for all $b\in\mathbb R^n$;
• Permutation of axis, e.g. $f(x)=Ax$ where $A$ is a permutation matrix.

Besides these trivial mapping (and their composition), I wonder if there are other distance-preserving mappings given a norm $\|\cdot\|$. As an example, if $\|\cdot\|$ is the ordinary Euclidean norm, then any orthogonal transformation satisfies $\|f(x)-f(z)\|=\|x-z\|$, which includes rotation, reflection, etc. However, for other norms, I cannot easily figure out such a non-trivial mapping. So my questions are:

• The most easiest question: for $\ell_p$ norm when $p\neq 2$, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|_p=\|x-z\|_p$? (I believe the answer is no despite the lack of a formal proof.)
• A more general question: if the answer of the first question if no, for any norm $\|\cdot\|$ which is not $\ell_2$, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|=\|x-z\|$?
• A further generalization: when using different norms for the input and the output, denoted as $\|\cdot\|_I$ and $\|\cdot\|_O$ respectively, does there exist a non-trivial $f$ such that $\|f(x)-f(z)\|_O=\|x-z\|_I$? In particular, I am interested in the case when $\|\cdot\|_I$ and $\|\cdot\|_O$ are $\ell_p$ norm and $\ell_q$ norm, respectively.