# Does an evolution family commute with the right shift semigroup?

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?)?

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.