Is every finite-order unimodular matrix conjugate to a $0,1,-1$ matrix? 
Problem. Given a matrix $A\in\mathrm{GL}(n,\mathbb{Z})$ such that $A^k=1$ for some $k\geq 1$, is there a matrix $g\in\mathrm{GL}(n,\mathbb{Z})$ such that $gAg^{-1}$ has only $0$, $1$, and $-1$ as possible entries?

Edit: after the remark by Mark Sapir that it is a famous open problem (which for me was already sufficient as an answer), I changed my question into the following ones, so now maybe it is more suitable for staying on MO without being closed.
What is known about this problem? Which other parts of mathematics is it connected to?
 A: I found a proof here for $n=4$: 
Yang, Qingjie, Conjugacy classes of torsion in (\mathrm{GL}_N(\mathbb Z)), Electron. J. Linear Algebra 30, 478-493 (2015). ZBL1329.15063. MR3414308
See the discussion in the last paragraph on p. 482 for the case that the characteristic polynomial is irreducible, and Theorem 1.7 for the reducible case. 
On the other hand, I suppose it's possible that the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ could grow faster than the number of $0,\pm1$ matrices intersected with $GL_n(\mathbb{Z})$. One can get a lower bound on the number of conjugacy classes of finite-order elements in $GL_n(\mathbb{Z})$ by counting the number which are block-diagonalizable with irreducible blocks. This should correspond to a sum over decompositions of $n$ into $\varphi(m)$ by $|Cl(\mathbb{Z}[e^{2\pi i/m}])|$, a sum over class numbers, since one obtains a conjugacy class of element of $GL_{\varphi(m)}(\mathbb{Z})$ of order $m$ for every ideal class in $\mathbb{Z}[e^{2\pi i/m}]$. I have no intuition though for the growth of this function, especially since the class numbers of cyclotomic fields behave erratically. 
A: For the record, the case $n=3$ of the problem can also be easily deduced by the lists presented in Tahara, On the finite subgroups of GL(3,Z)
