Let $\mathcal{S}(\mathbb{R})$ be the Schwartz-space, I define on it the two operators
$$\tau_h(f)(x) := f(x-h)$$ $$Df(x) := f'(x)$$
Now let $T : \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R})$ be an operator that commutes with $D$, is it true that $T$ commutes with $\tau_h$ for all $h$?
The heuristic idea behind this is the following
$$\tau_h(f)(x) = \sum_{k=0}^{\infty}{D^kf(x)\frac{h^k}{k!}}$$
So that If I apply $T$ both sides and commuting $T$ with $D^k$ I get
$$T\tau_hf = \sum_{k=0}^{\infty}{TD^kf\frac{h^k}{k!}} = \bigg[\sum_{k=0}^{\infty}{\frac{h^k}{k!}D^k}\bigg]Tf = \tau_hTf$$
Of course the problem is that is not true in general that $\sum_{k=0}^{\infty}{D^kf(x)\frac{h^k}{k!}} = f(x+h)$.
Actually the converse implication is really easy, just let $h \to 0$ in the equality
$$T(\frac{1}{h}(f - \tau_hf)) = \frac{1}{h}(Tf - \tau_hTf)$$