Isometry under condition $\|x-y\|=n\iff\|f(x)-f(y)\|=n.$ for $n\in\mathbf{N}$

Question

Let $X, Y$ be normed spaces and $f:X\to Y$ be mapping and $n\in\mathbf{N}$ If$$\|x-y\|=n\iff\|f(x)-f(y)\|=n.$$ Under what conditions this map will be an isometry?
Thanks

Answer 1

People define isometries differently - some specify that an isometry $f$ is always bijective. I will not assume this, but if you do, you need only add the condition "$f$ is surjective".

One sufficient condition would be that $f$ is a linear map. If this is the case, then for any $x,y\in X$, we have $$\left\|f\left(\frac{x}{\|f(x-y)\|}\right)-f\left(\frac{y}{\|f(x-y)\|}\right)\right\|=\left\|\frac{x}{\|f(x-y)\|}-\frac{y}{\|f(x-y)\|}\right\|$$ since the left hand side is equal to $1\in\mathbb{N}$. Thus, $\|f(x)-f(y)\|=\|x-y\|$.

If $f$ is not linear, I imagine we don't have enough information to give meaningful conditions for the function to be an isometry. If you replace $\mathbb{N}$ with $\mathbb{Q}$ then $f$ will be an isometry if and only if it is continuous.