Let $G_{\bullet} = (G_1 \rightrightarrows G_0)$ and $H_{\bullet} = (H_1 \rightrightarrows H_0)$ be Lie groupoids and $\varphi_\bullet, \psi_\bullet : G_\bullet \to H_\bullet$ be Lie groupoid morphisms, i.e. smooth functors. Suppose that $\eta : \varphi_\bullet \Rightarrow \psi_\bullet$ is a natural transformation, i.e. a smooth map $\eta : G_0 \to H_1$ such that $\eta(t(g))\varphi(g) = \psi(g) \eta(s(g))$ for all $g \in G_1$.

Thinking of $G_\bullet$ and $H_\bullet$ as representing stacks, this says that $\varphi_\bullet$ and $\psi_\bullet$ represent the same map $[G_0/G_1] \to [H_0/H_1]$. Hence (if my intuition is correct), there should be some kind of chain homotopy between the corresponding maps on the tangent complexes: $$ \begin{array}{ccccccc} 0 & \rightarrow & \operatorname{Lie}(G_\bullet) & \rightarrow & TG_0 & \rightarrow & 0\\ & &\varphi_1\downdownarrows\psi_1& \swarrow &\varphi_0\downdownarrows\psi_0 \\ 0 & \rightarrow & \operatorname{Lie}(H_\bullet) & \rightarrow & TH_0 & \rightarrow & 0. \end{array} $$ Is that correct, and if yes, how do we get the map $\swarrow$? One issue is that I'm not even sure what a chain homotopy would be in that context since these complexes of vector bundles do not live on the same manifolds. Is there a way to get a sort of chain homotopy capturing this idea? Note that differentiating $\eta$ doesn't automatically give us a map $TG_0 \to \operatorname{Lie}(H_\bullet)$.