Of course not. Take about any smooth $f(x,t)$ depending on $t$.

E.g., take $f(x,t)=x+t$, so that $$\phi^t_{af}(x_0)=\frac{a x_0 e^{a t}-a t+e^{a t}-1}{a}.$$

Of course not. Take about any smooth $f(x,t)$ depending on $t$.

E.g., take $f(x,t)=x+t$, so that for $a\ne0$ $$\phi^t_{af}(x_0)=\frac{a x_0 e^{a t}-a t+e^{a t}-1}{a},$$ so that $$\phi^t_{af}(x_0)-\phi^{at}_f(x_0)=\frac{(a-1) \left(1+a t-e^{a t}\right)}{a}\ne0$$ if $a\notin\{0,1\}$ and $t\ne0$.