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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 '12 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 '12 at 6:23
You seem to allow translations as well as permutations of coordinates, don't you? Do you treat the automorphism group as a subgroup of the isometries of the entire Hamming space? Or do you treat it as a subgroup of $Sym(S)$? The former is IMHO standard in coding theory (usually translations aren't allowed though). I am asking because the choice $S=\{111110^{n-5},0^n\}$ leads to a very large group with the "standard" interpretation: any permutation of the last $n-5$ coordinates $\times$ any permutation of the first five coordinates $\times$ two translations = $240(n-5)!$ automorphisms. – Jyrki Lahtonen Jun 10 '13 at 8:57

There's a better easy lower bound of $2^{n/3-1}(n/3)!$ and minimum distance six: start with the parity code of all $n/3$-bit words with an even number of ones, and replace the bits of each code word with $000$ or $111$.

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For the additive group {0, 1}^n , it seems to me that every non-singular binary nxn matrix provides one F_2 linear bijective map from {0, 1}^n to itself . As I recall, asymptotically the number of these bijective linear maps is at least C 2^(n^2), for some C>0 . In other words, a strictly positive proportion of random nxn matrices over F_2 has non-zero determinant, and an nxn matrix over F_2 has nxn entries, each being 0 or 1. When you say automorphism, are you referring to automorphisms of the additive group (F_2)^n ? Thanks. David Bernier

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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] – rishig Oct 19 '12 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. – David Bernier Oct 20 '12 at 17:47

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