Morphism of Lie groups $\theta:G\rightarrow H$ giving an equivalence of categories $BG\rightarrow BH$? Given a morphism of Lie groups $ \theta:G\rightarrow H$  and a principal $G$ bundle $ \pi:P\rightarrow M$ there are (at least) two ways to assign a principal $ H$ bundle.


*

*See that the morphism of Lie groups $  \theta:G\rightarrow H$ gives an action of $  G$ on $  H$ by $  g.h=\theta(g).h$. Given an action of $  G$ on manifold (Lie group in this case) $  H$ there is an associated fibre bundle $  P\times_G H\rightarrow M$ with fibre $  H$. This gives a principal $  H$ bundle.

*For principal bundle $  \pi:P\rightarrow M$, we can find an open cover $  \{U_\alpha\}$ of $  M$ and  (transition) maps $  g_\alpha g_\beta:U_{\alpha\beta}\rightarrow G$ satifsying the cocycle condition $  g_{\alpha\beta}g_{\beta\gamma}=g_{\alpha\gamma}$ on $  U_\alpha\cap U_\beta\cap U_\gamma$. Then the compositions $  \tau_{\alpha\beta}=\theta\circ g_{\alpha\beta}:U_{\alpha\beta}\rightarrow G\rightarrow H$ also satifies the cocycle condition $  \tau_{\alpha\beta}\tau_{\beta\gamma}=\tau_{\alpha\gamma}$ on $  U_\alpha\cap U_\beta\cap U_\gamma$. One can then produce a principal $  H$ bundle over $  M$ given this open cover $  \{U_\alpha\}$ of $  M$ and smooth maps $  \tau_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow H$ satisfying the cocycle condition. This gives a principal $  H$ bundle.


It is a good exercise (that I have not tried) to check that principal $  H$ bundles obtained from above two methods are (naturally) isomorphic.
Given a Lie group $  G$, let $  BG$ denote the category of principal $  G$ bundles. Objects are principal $  G$ bundles and morphisms are $  G$-equivariant morphisms.
Given a morphism of Lie groups $  \theta:G\rightarrow H$, above construction gives a functor (at the level of objects) $  B\theta:BG\rightarrow BH$. It is not difficult to see that, a $  G$-equivarint map induce a $ H$-equivariant map. This gives a functor.
I am trying to understand what can we say about $\theta:G\rightarrow H$ if we know that $B\theta:BG\rightarrow BH$ is an equivalence of categories? Does it have to be a diffeomorphism?  Any comments are welcome.
 A: It is a diffeomorphism, since there is an equivalence of bicategories $DifferentiableStacks \simeq LieGroupoids[W^{-1}]$ where the RHS is the bicategorical localisation of the usual 2-category of Lie groupoids a la Pronk. Because of the special nature of the domain of your $\theta$, namely it is a Lie groupoid with one object (call it $\mathbf{B}G$; note the boldface B!), then in fact 
$$
LieGroupoids[W^{-1}](\mathbf{B}G,\mathbf{B}H) \simeq LieGroupoids(\mathbf{B}G,\mathbf{B}H)
$$
is an equivalence of categories. The latter category is isomorphic to the category whose objects are homomorphisms $G\to H$ and whose 2-arrows are elements of $H$, acting by conjugation on homomorphisms. Tracing what happens to the quasi-inverse $\phi\colon BH\to BG$ to $B\theta$ through these equivalences of categories one gets a homomorphism $\psi\colon H\to G$ such that $\psi\circ \theta$ is conjugate to the identity map on $G$, and $\theta\circ \psi$ is conjugate to the identity map on $H$. This is enough to know that $\theta$ is a diffeomorphism, since we can pre- and post-compose $\psi$ with the inner automorphisms to get an inverse for $\theta$.
