$0$ is an stable equilibrium of $x' = Ax$ iff $A$ is semisimple, given that all of its eigenvalues have real part 0.
I'm kind of confused here: I had understood that if all of the eigenvalues of $A$ have real part 0 then the origin is an stable equilibrium of $x' = Ax$. That is because $A$ is a combination of rotations on different, linearly-independent subspaces.
What's wrong in my reasoning?
First of all, not all real matrices with purely imaginary eigenvalues are combinations of rotations; that is only true for some normal matrices. Also, if a normal matrix has a purely imaginary eigenvalue, it is a $90^\circ$ rotation followed by a scaling map.
More relevantly to the question at hand, not all real matrices with purely imaginary eigenvalues are diagonalizable (i.e. "semisimple"). As a counterexample, consider the matrix $$ A = \pmatrix{ 0&-1&1&0\\ 1&0&0&1\\ 0&0&0&-1\\ 0&0&1&0 } $$ Whose eigenvalues are $\pm i$. For systems with matrices like these, we need to study the behavior of generalized eigenvectors.