On page $10$ of the survey article *[Algebraic stacks](https://arxiv.org/abs/math/9911199)*, by T. Gomez (arXiv:math/9911199), we have following result



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> If a stack has an object with an automorphism other than the identity, then the stack can not be represented by a scheme.
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Is the converse true? 


Suppose I know only automorphism of any object are trivial automorphisms, then can I say that that stack is coming from a manifold/scheme. 


Is there any way to construct such scheme/manifold.


Is it practical to check if each object has only trivial automorphisms.


Edit : There is some slight confusion regarding the notion of "no non trivial automorphisms". I want to clear that (mostly to myself).


A stack over a category(site) $\mathcal{C}$ can be seen in two ways. 


One way as a category $\mathcal{D}$ with a functor $\mathcal{D}\rightarrow \mathcal{C}$ that is fibered in groupoids with some extra condition. Another way as a functor(almost) $F:\mathcal{C}^{op}\rightarrow (\text{Gpd})$ where $\text{Gpd}$ means the (2)category of groupoids.


To understand the notion of "no non trivial automorphisms" it is better (I think so) to see stack as $F:\mathcal{C}^{op}\rightarrow (\text{Gpd})$. By no non trivial automorphism for $F$ we mean the following :


Let $U$ be an object in $\mathcal{C}$. Then $F(U)$ is a groupoid. Let $x,y\in X(U)$ then, the set of arrows from $x,y$ denoted by $\text{Hom}(x,y)$ iseither empty or singleton. If this holds for each object $U$ of $\mathcal{C}$ and for each pair of objects $x,y\in F(U)$. Then, we say that this stack has no objects with non trivial automorphisms.


See a stack as a functor from category $\mathcal{D}$ to $\mathcal{C}$ say $F:\mathcal{D}\rightarrow \mathcal{C}$. Let $U$ be an object in $\mathcal{C}$. Look at the fiber category $\mathcal{D}(U)$. Let $x,y$ be objects in $\mathcal{D}(U)$ and $Hom(x,y)$ denote the arrows from $x,y$. If $Hom(x,y)$ is either singleton or empty for each pair of objects $x,y$ in $\mathcal{D}(U)$ and for each object $U$ in $\mathcal{C}$ then, we say the stack $\mathcal{D}\rightarrow \mathcal{C}$ has no non trivial automorphisms.



Stack being represented by a scheme/manifold
If all objects of a stack have trivial automorphism groups then it is equivalent to a sheaf, as pointed out by Daniel Litt in the comments. Pick your favourite non-representable sheaf as a counterexample to the claim that the stack is representable. For instance, on the site of manifolds, defined to be Hausdorff, paracompact, locally Euclidean topological spaces, consider the stack presented by a groupoid associated to an equivalence relation with non-Hausdorff quotient. In fact *any* equivalence relation such that the quotient is not a manifold will do (cf [the irrational flow on a torus](https://en.wikipedia.org/wiki/Linear_flow_on_the_torus#Irrational_winding_of_a_torus)). 


