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Here is one possible condition: If your normed space is complete with respect to the norm, and $f(x)$ is continuous and fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. For example, you can take $f(x)= \lambda x+ c$, with $\lambda, c\in R$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (check it)

2. it will be in fact true for all positive real number $n$, by taking the limit (the norm is continuous)

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$.

To sum up, assuming the space is complete, $f$ is an isometry if and only if $f$ is continuous, $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.

In general, I see no hope that there is much more to say about it : take for example the normed space to be the real numbers, and define $f(x) = {\rm floor}(x)$ (the largest integer just below $x$). Then $|x-y| = n$ if and only if $x = a+t$, with $a\in Z$ and $0\leq t<1$, $y=b+t$, with $b\in Z$, and $|a-b|=n$. So, if $|x-y|=n$, there holds obviously $|f(x)-f(y)| = n$, but it is clear that $f$ is not an isometry.

Notice also that if your relation holds for every rational number $n$ and the space is complete, then $f$ is an isometry by continuity.

Here is one possible condition: If $f(x)$ is continuous and fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (easy to check)

2. it will be in fact true for all positive real number $n$ : recall that the norm is continuous and take the limit of a sequence $(x_m,y_m)$ such that $x_m\to x$, $y_m\to y$, $||x_m-y_m||\in \mathbb Q$, and $||x_m-y_m||\to ||x-y||$ (such a sequence exists because the subspace spanned by $x,y$ is isometric to $\mathbb R^2$, endowed with the transported norm, equivalent to any other norm in $\mathbb R^2$).

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$.

To sum up, $f$ is an isometry if and only if $f$ is continuous, $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.

Notice also that if your relation holds for every rational number $n$ and $f$ is continuous, then $f$ is an isometry by continuity.

Here is one possible condition: If your normed space is complete with respect to the norm, and $f(x)$ fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. For example, you can take $f(x)= \lambda x+ c$, with $\lambda, c\in R$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (check it)

2. it will be in fact true for all positive real number $n$, by taking the limit (the norm is continuous)

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$. To sum up, assuming the space is complete, $f$ is an isometry if and only if $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.

In general, I see no hope that there is much more to say about it : take for example the normed space to be the real numbers, and define $f(x) = {\rm floor}(x)$ (the largest integer just below $x$). Then $|x-y| = n$ if and only if $x = a+t$, with $a\in Z$ and $0\leq t<1$, $y=b+t$, with $b\in Z$, and $|a-b|=n$. So, if $|x-y|=n$, there holds obviously $|f(x)-f(y)| = n$, but it is clear that $f$ is not an isometry.

Notice also that if your relation holds for every rational number $n$ and the space is complete, then $f$ is an isometry by continuity.

Here is one possible condition: If your normed space is complete with respect to the norm, and $f(x)$ is continuous and fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. For example, you can take $f(x)= \lambda x+ c$, with $\lambda, c\in R$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (check it)

2. it will be in fact true for all positive real number $n$, by taking the limit (the norm is continuous)

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$. 

To sum up, assuming the space is complete, $f$ is an isometry if and only if $f$ is continuous, $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.

In general, I see no hope that there is much more to say about it : take for example the normed space to be the real numbers, and define $f(x) = {\rm floor}(x)$ (the largest integer just below $x$). Then $|x-y| = n$ if and only if $x = a+t$, with $a\in Z$ and $0\leq t<1$, $y=b+t$, with $b\in Z$, and $|a-b|=n$. So, if $|x-y|=n$, there holds obviously $|f(x)-f(y)| = n$, but it is clear that $f$ is not an isometry.

Notice also that if your relation holds for every rational number $n$ and the space is complete, then $f$ is an isometry by continuity.

Here is one possible condition: If your normed space is complete with respect to the norm, and $f(x)$ fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. For example, you can take $f(x)= \lambda x+ c$, with $\lambda, c\in R$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (check it)

2. it will be in fact true for all positive real number $n$, by taking the limit (the norm is continuous)

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$.

In general, I see no hope that there is much more to say about it : take for example the normed space to be the real numbers, and define $f(x) = {\rm floor}(x)$ (the largest integer just below $x$). Then $|x-y| = n$ if and only if $x = a+t$, with $a\in Z$ and $0\leq t<1$, $y=b+t$, with $b\in Z$, and $|a-b|=n$. So, if $|x-y|=n$, there holds obviously $|f(x)-f(y)| = n$, but it is clear that $f$ is not an isometry.

Notice also that if your relation holds for every rational number $n$ and the space is complete, then $f$ is an isometry by continuity.

Here is one possible condition: If your normed space is complete with respect to the norm, and $f(x)$ fulfills $||f(kx)-f(ky)|| = k||f(x)-f(y)||$, for every $k \in N$. For example, you can take $f(x)= \lambda x+ c$, with $\lambda, c\in R$. This works because :

1. the equivalence you wrote above will be true not only for $n\in N$ but also for every positive rational number $n$ (check it)

2. it will be in fact true for all positive real number $n$, by taking the limit (the norm is continuous)

You can even restrict this condition to hold only whenever $||kx - ky||$ is a natural number, and not for all $x,y$. To sum up, assuming the space is complete, $f$ is an isometry if and only if $||x-y|| = n$ implies $||f(x)-f(y)||=n$ for all $n\in N$ and $||f(kx)-f(ky)|| = k ||f(x)-f(y)||$ holds for all $k\in N$.

In general, I see no hope that there is much more to say about it : take for example the normed space to be the real numbers, and define $f(x) = {\rm floor}(x)$ (the largest integer just below $x$). Then $|x-y| = n$ if and only if $x = a+t$, with $a\in Z$ and $0\leq t<1$, $y=b+t$, with $b\in Z$, and $|a-b|=n$. So, if $|x-y|=n$, there holds obviously $|f(x)-f(y)| = n$, but it is clear that $f$ is not an isometry.

Notice also that if your relation holds for every rational number $n$ and the space is complete, then $f$ is an isometry by continuity.