Show that if $u(t)$ solves $\dot{u} = Au$, then $v(t) = u(-t)$, solves $\dot{v} = Bv$, where $B = -A$.
Similarly, show that if u(t) solves $\dot{u} = Au$, then $v(t) = u(2t)$ solves $\dot{v} = Bv$, where $B = 2A$.
I don't know how to formally prove this. It seems obvious because since the solution $v(t) = u(-t)$ is the same as $u(t)$ but it has a scalar which is negative and a scalar can be taken out of the solution and placed outside which follows that we have $-A$. The same goes with the second part of the question.