Let $n$ be a positive integer. It is easy to find a symmetric matrix with complex coefficients with order $n$ i.e. $A^n = I$, one can just consider the diagonal matrix with $n$-th roots of units.
A matrix $A$ has order $n$ if it is the smallest positive integer such that $A^n = I$.
But is there a symmetric matrix with integer coefficients with order $n$?
Hint Since $A^n = {\bf 1}$, all of its eigenvalues are $n$th roots of unity. But all of the eigenvalues of a real symmetric matrix are real. (It could be helpful, too, to remember that symmetric matrices are diagonalizable.)