## Automorphisms of subgroup of hamming cube under distance constraint

Let $S$ be a subset of $\{0,1\}^n$ such that any two elements of $S$ are at least (Hamming) distance 5 apart. I'm looking for an upper bound on the size of the automorphism group of $S$.

There's a trivial upper bound of $2^nn!$ (the number of automorphisms of $\{0,1\}^n$), and an easy lower bound of $2^{n/5}(n/5)!$ (take S to be all elements of the form $xxxxx$, where $x$ is a bitstring of length $n/5$).

Any bound of the form $n!/n^{cn}$ for any $c>0$ would be helpful.

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 Such a set would be a 2-error correcting binary code. You might search for constructions and see if some provide good lower bounds. – Aaron Meyerowitz Oct 14 at 22:00 Thanks for the comment. Haven't found anything yet that I've gotten to work, but I'll post here if I do. – rishig Oct 19 at 6:23

 Thanks for your response. I'm referring to automorphisms of the hypercube [1], which are substantially more restricted than $F_2^n$. For instance the linear map [[1 1] [1 0]] is an automorphism of $F_2^2$ but not of the square. [1] en.wikipedia.org/wiki/Hyperoctahedral_group – rishig Oct 19 at 4:44 I might be mistaken. But perhaps the autmorphism group is the same as the hroups that preserve the Hamming distance on the metric space {0,1}^n with the Hamming distance metric? Noam Elkies wrote a survey article on linear error-correcting codes that was published in the Notices of the AMS: Elkies, Noam; "Lattices, Linear Codes, and Invariants, Part II", Notices of the AMS, Volume 47 number 11, December 2000. ams.org/notices/200011/index.html – David Bernier Oct 20 at 17:47