Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. We have a representation of $\mathfrak{g}$ on $C^\infty(G)$ via interpreting elements of the Lie algebra as left-invariant derivations, or equivalently $$(Xf)(g) = \frac{d}{dt}f(g\exp(tx))\big|_{t = 0},$$ where $f \in C^\infty(g), X \in \mathfrak{g}$, and $g \in G$. This action extends to the universal enveloping algebra $U\mathfrak{g}$ via the universal property; explicitly, a monomial $X_1 \dots X_n \in U\mathfrak{g}$ acts by $$(X_1 \dots X_n f)(g) = \frac{\partial^n}{\partial t_1 \dots \partial t_n} f(g\exp(t_1X_1) \dots \exp(t_nX_n))\big|_{t_i = 0 \forall i}.$$
We also have the right regular action $\rho: G \curvearrowright C^\infty(G)$ via $\rho_h(f)(g) = f(gh)$. Exercise 10.2(ii) in Bump's Lie Groups asks you to show that if $D$ lies in the center of $U\mathfrak{g}$, then the actions of $D$ and the right regular action commute. (The left regular action always commutes with the action of $U \mathfrak{g}$.)
Since we assume $G$ connected, it suffices to show that $\rho_h$ commutes with $D$ when $h$ is of the form $\exp(X)$ for $X \in \mathfrak{g}$. Intuitively, I want to say that $$((\rho_{\exp(X)} \circ D)(f))(g) = \frac{d}{dt}f(g \exp(X) \exp(tD))\Big|_{t = 0},$$ use centrality of $D$ to commute the two exponentials, and conclude. However, the expression $\exp(tD)$ is nonsensical for an arbitrary element $D \in U \mathfrak{g}$, and I don't see how to make progress using the explicit expression for the action written above. Does anyone have any hints on how to make this argument rigorous?
$G$ acts on $\mathfrak g$ by the adjoint action, which extends to an algebra automorphism of $U(\mathfrak g)$.
By Lemma 23.2 of Humphreys, for any central element $D\in U(\mathfrak g)$, the adjoint action is trivial: $\mathrm{Ad}(g)D=D$. Thus, it suffices to prove the following:
Lemma For any $X\in U(\mathfrak g)$, we have $\rho_g\circ X\circ \rho_g^{-1}=\mathrm{Ad}(g)X$.
But to prove the lemma, since both $\mathrm{Ad}(g)$ and $\rho_g\circ -\circ \rho_g^{-1}$ respect multiplication and addition, it suffices to check it on the generators, i.e., on elements $X\in\mathfrak g$. But then this is clear.