Is the right-hand term of the autonomous dynamic system equivalent to the original system after being multiplied by a constant?

Question

Given two dynamical systems where $f$ is lipschitz for $x$ : $\begin{cases} x'(t)=af(x),\\ x(0)=x_0,\end{cases} t\in[0,\tau]$ and $\begin{cases} z'(t)=f(z),\\ z(0)=x_0,\end{cases} t\in[0,\tau']$, and their flowmaps $\phi_{af}^\tau(x),\phi_f^{\tau'}(x)$, if $\tau'=a\tau$, can we say $\phi_{af}^\tau(x_0)=\phi_f^{\tau'}(x_0)$ for any $x_0 \in \mathbb{R}$?

How about in a higher dimension? $\begin{cases} x'(t)=Af(x),\\ x(0)=x_0,\end{cases} t\in[0,\tau]$ and $\begin{cases}z'(t)=f(z),\\ z(0)=x_0,\end{cases} t\in[0,\tau']$ where $A\in\mathbb{R}^{n\times n}$ is diagonal matrix? what is the relationship between $\phi_{Af}^\tau(x_0)$ and $\phi_f^{\tau'}(x_0)$

This is for autonomous systems, the same question on non-autonomous systems has been solved in

Is the right-hand term of the dynamic system equivalent to the original system after being multiplied by a constant?

Answer 1

In the autonomous case the answer is yes.

Indeed, suppose that $$\text{$z'(t)=f(z(t))$ for $t\in[0,\tau']=[0,a\tau]$ and $z(0)=x_0$.}$$ For $t\in[0,\tau]$, let $X(t):=z(at)$. Then $$\text{$X'(t)=az'(at)=af(z(at))=af(X(t))$ for $t\in[0,\tau]$ and $X(0)=x_0$.}$$ So, $X(\cdot)$ is a solution $x(\cdot)$ to the initial value problem $$\text{$x'(t)=af(x(t))$ for $t\in[0,\tau]$ and $x(0)=x_0$.}$$ In view of the Lipschitz condition, this solution is unique. So, $$\phi_{af}^\tau(x_0)=X(\tau)=z(a\tau)=z(\tau')=\phi_f^{\tau'}(x_0).\quad\Box$$