The answer is "no". Consider a smooth, proper $1$-dimensional Deligne-Mumford stack over $\mathbb{C}$ which has coarse moduli space $\mathbb{P}^1$ and which has a single "stacky" point (with any nontrivial stabilizer group at that point).

**Edit.** A more precise definition of the stack is as follows. Let $m$ be an integer, $m>1$. Let $A$ be $\mathbb{A}^2 \setminus \{(0,0)\}$ with coordinates $x$ and $y$. Let $\mathbb{G}_m$ act on $A$ by $t\ast (x,y) = (tx,t^m y)$. The stack $Q_m$ is the quotient stack $[A/\mathbb{G}_m]$. The quotient morphism $A\to Q_m$ is a smooth atlas, implying that $Q_m$ is smooth (since $A$ is smooth). In fact, this morphism is naturally a $\mathbb{G}_m$-torsor; denote the associated invertible sheaf on $Q_m$ by $\mathcal{L}$. It is not hard to see that this is an $m$-torsion invertible sheaf; the $\text{m}^{\text{th}}$ tensor power of $A$ as a $\mathbb{G}_m$-torsor over $Q_m$ is the quotient stack $[(\mathbb{G}_m\times A)/\mathbb{G}_m]$ where the $\mathbb{G}_m$-action is $t\cdot (u,(x,y)) = (t^mu, (tx,t^my))$. This admits a "trivializing" morphism to $\mathbb{G}_m$, $(u,(x,y)) \mapsto u/x^m = u/y$, each defined on the appropriate open $\mathbb{G}_m\times D(X)$ or $\mathbb{G}_m\times D(y)$. The only nontrivial stabilizer group is $\mathbf{\mu}_m$ acting on the $\mathbb{G}_m$-orbit $\{0\}\times \mathbb{G}_m$ inside $A$. So $Q_m$ is a Deligne-Mumford stack. Finally, $Q_m$ is finite over its coarse moduli space $\mathbb{P}^1$, where the $\mathbb{G}_m$-invariant morphism $A\to \mathbb{P}^1$ is just $(x,y) \mapsto [x^m,y]$.

truein the differentiable category; the result is in the thesis of Dorette Pronk. – David Carchedi Feb 3 '13 at 22:10