Let $X$ be a Banach space and
$$\mathrm C\_0(\mathbb R,X):=\{f\colon\mathbb R\to X\colon f \text{ is continuous and } \lim\limits\_{|t|\to\infty}f(t)=0\}$$
normed by
$$\|f\|:=\sup\limits\_{t\in\mathbb R}\|f(t)\|, \quad f\in \mathrm C\_0(\mathbb R,X).$$
Let $(U(t,s))\_{t\ge s}$ denote an evolution family of bounded linear operators on $X$ satisfying:
$$U(t,r)U(r,s)=U(t,s)$$
and $U(t,t)=\operatorname{Id}$, the identity on $X$, for all real numbers $s\le r\le t$. Let $(R(t))\_{t\ge 0}$ denote the right shift semigroup:
$$(R(t)f)(s)=f(t-s)$$
for all $s\in\mathbb R$, $t\ge 0$, and $f\in \mathrm C\_0(\mathbb R,X)$.


Does the evolution family and the right shift semigroup commute? More precisely, it is true that for any $T\ge 0$
$$U(t,t-s)R(s)f=R(s)U(t,t-s)f$$
holds for all $t,s\in[0,T]$ and $f$ from some appropriate space (which?)?



Does an evolution family commute with the right shift semigroup?
In general no; this happens only in very special cases.


To make this simpler, let's suppose $U$ is a semigroup (a time-homogeneous evolution family), so $U(s,t) =U(t-s)$. Then formally, this commutation would only occur if the generator $A$ of $U(t)$ were to commute with the generator of $R(t)$, which is $\frac{d}{dx}$. So we get counterexamples by taking $U(t)$ to be generated by any operator $A$ which does not commute with $\frac{d}{dx}$. 


Perhaps the simplest counterexample is to let $A$ be multiplication by any nonnegative non-constant continuous function $h$, so $U(t)f = e^{-th} f$. Then you can easily see that $$\begin{align\*}(U(t)R(s)f)(x) &= e^{-th(x)} f(x-s) \\ (R(s) U(t) f)(x) &= e^{-th(x-s)} f(x-s)\end{align\*}$$ which are not the same.


