Given a positive integer $n$, the [Hamming distance](https://en.wikipedia.org/wiki/Hamming_distance) $d^H\_n(x,y)$ of $x,y\in \{0,1\}^n$ is defined by $$d^H\_n(x,y) = |\{k\in\{0,\ldots,n-1\}: x(k)\neq y(k)\}|.$$ 


Given an integer $n>0$ and a set $S\subseteq \{0,1\}^n$ with $|S| = n$, is it possible to find a map $f:S\to \{0,1\}^{n+1}$ such that $$d^H\_{n+1}(f(x), f(y)) = d^H\_n(x,y) + 1 \text{ for all } x\neq y\in S$$?



Stretching map of $n$ points from $\{0,1\}^n$ to $\{0,1\}^{n+1}$ with respect to their Hamming distance
Here is an infinite family of counterexamples. Let $n = 2^r$. The $2^r$ distinct rows of a $2^r \times 2^r$ Hadamard matrix with entries from $\{0,1\}$ are all at distance $2^{r-1}$. However the maximum number of binary words all at distance $2^{r-1} + 1$ in $\{0,1\}^{2^r+1}$ is just $2$. *Proof.* We can suppose that $0\ldots00 \ldots 0$ and $1\ldots 11\ldots 0$ are two of these words. Flipping any $2^{r-1} + 1$ bits in the second word gives a binary word of even weight, so not at the odd distance $2^{r-1}+1$ from all-zeros.


