Let $N=(1,0,0)$ denote the North Pole of $\mathbb S^2$. Therefore, $-N$ denotes the South Pole.
Question. Is there a skew-symmetric $M\in\mathbb R^{3\times 3}$ such that, letting $u(t,P)\in\mathbb S^2$ denote the solution to $$\frac{\mathrm{d}u}{\mathrm{d}t}= Mu,\qquad u(0)=P,$$ for an arbitrary $P\in\mathbb S^2$, it holds that either $u(t,P)\to N$ or $u(t,P)\to -N$ as $t\to \infty$?
The matrix $M$ must satisfy the skew-symmetry $M^T=-M$ in order for $u(t, P)$ to remain on $\mathbb S^2$ for all $t>0$. In particular, the trace of $M$ must vanish.
My intuition is that such a dynamical system cannot exist. Indeed, since $\operatorname*{tr}(M)=0$, the dynamical system $u(t, \cdot)$ should be "area-preserving", in some sense. And this shouldn't be compatible with the requirement that, asymptotically, everything collapses to a couple of points, which is a set with zero area.
However, if my intuition is wrong, I would be very happy to see an example.
You can solve the system explicitly to see that this isn't possible. Since $M$ is skew-symmetric, up to an orthogonal change of coordinates, it looks like
$$ M = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & a \\ 0 & -a & 0 \end{pmatrix}. $$
If you start with $P = \pm e_1$ then the corresponding dynamical system will just stay at $P$ (as $Me_3 = 0$). However, if you start at (say) $e_2$, the initial value problem is
$$ \begin{pmatrix} u_1'(t) \\ u_2'(t) \\ u_3'(t) \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & a \\ 0 & -a & 0 \end{pmatrix} \begin{pmatrix} u_1(t) \\ u_2(t) \\ u_3(t) \end{pmatrix} = \begin{pmatrix} 0 \\ au_3(t) \\ -au_2(t) \end{pmatrix}, \,\,\, \begin{pmatrix} u_1(0) \\ u_2(0) \\ u_3(0) \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} $$
whose solution is
$$ u_1(t) \equiv 0, u_2(t) = \cos(at), u_3(t) = -\sin(at). $$
The trajectory is a big circle on the $yz$ plane which always stays away from $\pm e_1$.