I have encountered the problem:
Prove that if $\phi_t(x)$ is the flow of $\dot x=f(x)$ then the function $\Phi_t(x)=e^{at}\phi_t(x)$ is the flow of $\dot x = axf(x)$.
I do not think this statement is true, and have no idea how to prove it. Trying it for equations $\dot x=-x$ and $\dot x = - ax^2$ has resulted in the statement being false. Can someone provide any hints or point me in the right direction? Is it a typo in the problem setting maybe?
By the product rule $$ \frac{d}{dt} Φ_t(x)=ae^{at}ϕ_t(x)+e^{at}\frac{d}{dt}ϕ_t(x) =aΦ_t(x)+e^{at}f(e^{-at}Φ_t(x)) $$ which seems far away from the claim.
From the other end, the claim only makes sense for scalar ODE, then the second ODE can be simplified to $\frac{d}{dt} \ln x=af(x)$ which does not have an easy connection to the first ODE.
The only typo that might be close to the problem is if $Φ_t(x)=e^{aϕ_t(x)}$ was intended. Then $$ \frac{d}{dt} Φ_t(x)=ae^{atϕ_t(x)}\frac{d}{dt}ϕ_t(x) =aΦ_t(x)f(a^{-1}\ln(Φ_t(x))) $$ which still does not result in something similar.
In the end it might really be the first variant. Let $A$ be a matrix, $e^{At}$ the matrix exponential, then the modified system is $\dot x=Ax+f_A(t,x)$ where $f_A(t,x)=e^{At}f(e^{-At}x)$. This is still not very close, but useful in the case where the function $f$ is split in a dominant linear part and a locally small superlinear part. See operator-splitting methods, exponential one-step methods, etc.