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Motivated by this question, what can be said about the number $f(n)$ of adjoint orbits of $\mathrm{Mat}(n,\mathbb{C})$ (the ring of all $n\times n$ complex matrices) that contain a $(0,1)$-matrix? More concretely, define two matrices $A,B\in \mathrm{Mat}(n,\mathbb{C})$ to be equivalent if there is a matrix $X\in \mathrm{GL}(n,\mathbb{C})$ such that $A=XBX^{-1}$. How many equivalence classes contain a $(0,1)$-matrix? I doubt whether an exact answer is feasible, but are there good estimates or an asymptotic formula? For instance, is it true that $f(n)=2^{n^2+o(n^2)}$?

Numerous variations are possible, such as replacing $\mathbb{C}$ with some other field (or even a ring like $\mathbb{Z}$) and replacing $(0,1)$-matrices with other classes of matrices, such as matrices with integer entries between integers $a$ and $b$, with $a<b$.

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  • $\begingroup$ Perhaps simpler question as a starting point might be: what's the probability that two (uniformly random) $(0,1)$-matrices are equivalent? $\endgroup$
    – Sam Hopkins
    Nov 30, 2019 at 1:28
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    $\begingroup$ For small values of n, Smith Normal Form ( of order n 0-1 matrices) has been enumerated. Does this lend a clue? I suspect number of equivalence classes will be much smaller, closer to (but not as small as) number of distinct determinants. Gerhard "Will Look Up Zivkovic's Paper" Paseman, 2019.11.29. $\endgroup$
    – Gerhard Paseman
    Nov 30, 2019 at 1:58
  • $\begingroup$ @GerhardPaseman: There is little connection (except for the determinant) between SNF over $\mathbb{Z}$ and adjoint orbits over $\mathbb{C}$. For instance, any two $n\times n$ matrices over $\mathbb{Z}$ with equal squarefree determinants have the same SNF over $\mathbb{Z}$. On the other hand, the matrices $\left[ \begin{array}{cc} 2 & 2\\ 0 & 2\end{array}\right]$ and $\left[ \begin{array}{cc} 2 & 1\\ 0 & 2\end{array}\right]$ have different SNF's over $\mathbb{Z}$ but lie in the same adjoint orbit over $\mathbb{C}$. $\endgroup$
    – Richard Stanley
    Nov 30, 2019 at 16:45
  • $\begingroup$ @Richard, If you change the question to over Z, then I agree. If you are keeping it in the realm of 0-1 matrices, I would like to see a similar example to what you provided, but using 0-1 matrices instead. Gerhard "You'll Need A Bigger Order" Paseman, 2019.11.30. $\endgroup$
    – Gerhard Paseman
    Nov 30, 2019 at 23:53
  • $\begingroup$ @GerhardPaseman: Let $A=\left[ \begin{array}{cccccc} 0 & 1 & 1 & 0 & 1 & 1\\ 1 & 0 & 1 & 1 & 0 & 1\\ 1 & 1 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 1 & 0\end{array}\right]$ and $B=\left[ \begin{array}{cccccc} 0 & 1 & 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 & 0 & 1\\ 0 & 0 & 0 & 1 & 1 & 0\end{array}\right]$. Then $A$ and $B$ have different SNF's over $\mathbb{Z}$ but lie in the same adjoint orbit over $\mathbb{C}$. $\endgroup$
    – Richard Stanley
    Dec 1, 2019 at 3:01

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