Given a manifold $M$, the collection of all automorphisms of $M$, denoted by $\text{Aut}(M)$ forms a Lie group.


Do we have similar setting in case of Lie groupoid?


Is there "a structure" whose "automorphisms" forms a Lie groupoid?



Automorphisms of which structure form a Lie groupoid
Here is a construction due to Ehresmann, and covered in detail by Mackenzie in either of his Lie groupoids books. 


Take a principal $G$-bundle, $\pi\colon P\to M$, everything here in smooth manifolds. Then there is a Lie groupoid with object manifold $M$ and and a morphism from $m\_1$ to $m\_2$ is a triple $(m\_1,m\_2,\phi)$ where $\phi$ is an isomorphism $\phi\colon P\_{m\_1} \stackrel{\simeq}{\to} P\_{m\_2}$. This collection of morphisms has the structure of a smooth manifold, and the functions $(m\_1,m\_2,\phi) \mapsto m\_i$ are surjective submersions. Composition is smooth, and inversion is $(m\_1,m\_2,\phi)\mapsto (m\_2,m\_1,\phi^{-1})$. This gives a Lie groupoid.


In slightly more detail, the collection of all isomorphisms $P\_{m\_1} \stackrel{\simeq}{\to} P\_{m\_2}$ forms a manifold diffeomorphic to the underlying manifold of $G$. The only hard part is to see how these are collected up into a single manifold. One can look locally around $m\_1$ and $m\_2$, where $P$ is trivial, and then you are looking at isomorphisms of trivial bundles, which are essentially parametrised collections of automorphisms of $G$ as a $G$-space (so, not as a group). The arrow manifold of this Lie groupoid is isomorphic to the quotient of $P\times P$ by the diagonal action of $G$, namely $(p\_1,p\_2) \mapsto (p\_1g,p\_2g)$. The source and target maps are given by $[p\_1,p\_2]\mapsto \pi(p\_i)$. (**Edited** to confirm and correct this definition).




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The automorphisms $X\stackrel{\simeq}{\to} X$ of a single object $X$ will *never* give rise to a Lie groupoid that has more than one object, since you somehow need to break being able to always compose automorphisms.


