Complexity to calculate a full set of eigenvectors over a finite field
There will generally not exist a set of $n$ linearly independent vectors $x$ such that $Ax = x$; this can only happen for $A$ being the identity matrix. On the other extreme, there may be no such vectors at all. For example, the matrix
$$
A = \left(\begin{array}{cc} 0 &amp; 1\\ 1 &amp; 1 \end{array}\right)
$$
has full rank, but there are no nonzero vectors $x$ such that $Ax = x$. This is consistent with the fact that the characteristic polynomial of $A$ is $\lambda^2 + \lambda + 1$, which is irreducible over GF(2).

If you want to find the maximum number $k$ of linearly independent vectors $x$ such that $Ax=x$, compute the nullspace of $A - I$. This can be done using Gaussian elimination in time $O(n^3)$. Several other algorithms for this task also exist, and are described in textbooks on computational linear algebra.