= Solution
{c}
Using $x=-u$, $y=-v$ in the original <autonomous planar system> gives
$$
\boxed{\dot u=v+1,\qquad \dot v=1-u^2.}
$$
The map $(x,y)\mapsto(-x,-y)$ is a <rotation> by $\pi$, preserving time and the clockwise or counterclockwise orientation of trajectories. Hence the given <centre equilibrium> at $(-1,1)$ becomes a \b[clockwise centre at $(1,-1)$], and the given <saddle equilibrium> at $(1,1)$ becomes a \b[saddle at $(-1,-1)$]. These are the only <equilibrium points>, because <rotation> is invertible.
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