Solutions of autonomous system $\dot{x} = f(x)$ if $f\circ T = -T\circ f$ for some nonsingular matrix $T$ Having an autonomous system $\dot{x} = f(x)$ with general solution $\phi(t, \xi)$. If $T$ is an $m \times m$ nonsingular matrix such that $f(Tx) = -Tf(x)$ for all $x\in \mathbb{R}^m$ prove $\phi(t, T\xi) \equiv T\phi(-t, \xi)$
I can check easily that $\phi(t, T\xi) \equiv T\phi(-t, \xi)$ for $t=0$. Then I tried to check that their derivatives over time are equal (using the properties of $T$ and $f(x)$) but I got stuck.
I also tried considering two systems, $\dot{x} = f(x)$ and $\dot{y} = -f(y)$ where $y = Tx$, perform the change of variable with the general solution and try to get the equality without luck.
I am using the right approach? Can anyone give a hint?
Thanks
 A: Fix $\xi\,$, and consider the function $$\psi(t):=T\phi(-t,\xi)\ .$$ From 
$$\dot\psi(t)=-T\dot\phi(-t,\xi)=-Tf\bigl(\phi(-t,\xi)\bigr)=f\bigl(T\phi(-t,\xi)\bigr)=f\bigl(\psi(t)\bigr)$$
it follows that $\psi$ is a solution of the given system, and as $$\psi(0)=T\phi(0,\xi)=T\xi$$
we can safely say that in fact
$$\psi(t)=\phi(t,T\xi)\ .$$
A: You're following the right approach, but you're missing one thing: the interpretation of the second component, $\xi$, of the flow $\phi(t,\xi)$. To get you started:
Take the system $\dot{x} = f(x)$. Denote its flow by $\phi$, then the value of $x$ after time $t$, for the initial condition $x(0) = \xi$, is given by $x(t) = \phi(t,\xi)$. 
Now consider the system $\dot{y} = g(y)$. Denote its flow by $\psi$, then the value of $y$ after time $t$, for the initial condition $y(0) = \eta$, is given by $y(t) = \psi(t,\eta)$. 
Now, what happens if you add the information that 1) $y = T x$ and 2) $g(y) = - f(y)$? In particular, how can you use 1) to transform $\eta$?
