Maybe It is time to sum up the work that has been down up to now :

Let $X$ and $Y$ be complete normed spaces, and $f$ a function $X\to Y$. Denote by $\cal C$ the condition "$||x-y|| = n\in \mathbb N \Rightarrow ||f(x)-f(y)|| = n$".

1. If $\dim(X)=1$, no proper condition can be added to $\cal C$, and even to the stronger condition set by the asker, that would imply that $f$ be an isometry (MikeTex) ;

2. If $\dim(X) > 1$, then $f$ is an isometry if and only if $f$ is continuous and fulfills the condition "$||f(kx)-f(ky)|| = k||f(x)-f(y)||$ for every $x,y$, and every $k\in \mathbb N$" (MikeTex) ;

3. If $X$ and $Y$ are Hilbert space and $\dim(X)>1$, then condition $\cal C$ implies that $f$ is an isometry (Terry Tao) ;

4. More generally, if $\dim(X)>1$ and $Y$ is strictly convex ($X$ and $Y$ being, as in the hypotheses, complete with respect to the norms), then condition $\cal C$ implies that $f$ is an isometry (Eric Wofsey).

It is interresting that the additional condition $||f(x)-f(y)|| = n \Rightarrow ||x-y|| = n$ present in the question of the asker has not been used in the argument of Eric Wofsey (hence is unnecessary, if used, in the argument of Terry Tao). The question is now : can this additional condition be used in order to weaken the assumptions of Eric and Terry.

Maybe It is time to sum up the work that has been down up to now :

Let $X$ and $Y$ be normed spaces, and $f$ a function $X\to Y$. Denote by $\cal C$ the condition "$||x-y|| = n\in \mathbb N \Rightarrow ||f(x)-f(y)|| = n$".

1. If $\dim(X)=1$, no proper condition can be added to $\cal C$, and even to the stronger condition set by the asker, that would imply that $f$ be an isometry (MikeTex) ;

2. If $\dim(X) > 1$, then $f$ is an isometry if and only if $f$ is continuous and fulfills the condition "$||f(kx)-f(ky)|| = k||f(x)-f(y)||$ for every $x,y$, and every $k\in \mathbb N$" (MikeTex) ;

3. If $X$ and $Y$ are Hilbert space and $\dim(X)>1$, then condition $\cal C$ implies that $f$ is an isometry (Terry Tao) ;

4. More generally, if $\dim(X)>1$ and $Y$ is strictly convex, then condition $\cal C$ implies that $f$ is an isometry (Eric Wofsey).

It is interresting that the additional condition $||f(x)-f(y)|| = n \Rightarrow ||x-y|| = n$ present in the question of the asker has not been used in the argument of Eric Wofsey (hence is unnecessary, if used, in the argument of Terry Tao). The question is now : can this additional condition be used in order to weaken the assumptions of Eric and Terry.