ag.algebraic geometry - Are root stacks characterized by their divisor multiplicities?

Definitions/Background

Suppose $S$ is a scheme and $Dsubseteq S$ is an irreducible effective Cartier divisor. Then $D$ induces a morphism from $S$ to the stack $[mathbb A^1/mathbb G_m]$ (a morphism to this stack is the data of a line bundle and a global section of the line bundle, modulo scaling). For a positive integer $k$, the root stack $sqrt[k]{D/S}$ is defined as the fiber product

$begin{matrix}
sqrt[k]{D/S} & longrightarrow & [mathbb A^1/mathbb G_m] \
pdownarrow & & downarrow wedge k \
S & longrightarrow & [mathbb A^1/mathbb G_m]
end{matrix}$

where the map $wedge k: [mathbb A^1/mathbb G_m]to [mathbb A^1/mathbb G_m]$ is induced by the maps $xmapsto x^k$ (on $mathbb A^1$) and $tmapsto t^k$ (on $mathbb G_m$). The morphism $p:sqrt[k]{D/S}to S$ is a coarse moduli space and is an isomorphism over $Ssmallsetminus D$. Moreover, there is a divisor $D'$ on $sqrt[k]{D/S}$ such that $p^*D$ is $kD'$.

The data of a morphism from $T$ to $sqrt[k]{D/S}$ is equivalent to the data a morphism $f:Tto S$ and a divisor $E$ on $T$ such that $f^*D = kE$.

The question

Suppose $mathcal X$ is a DM stack, that $f:mathcal Xto S$ is a coarse moduli space, that $f$ is an isomorphism over $Ssmallsetminus D$, and that $f^*D = kE$ for an irreducible Cartier divisor $E$ on $mathcal X$. Is the induced morphism $mathcal Xto sqrt[k]{D/S}$ an isomorphism?

I get the strong impression that the answer should be "yes", at least if additional conditions are placed on $mathcal X$.

A counterexample

Here's a counterexample to show that some additional condition needs to be put on $mathcal X$. Take $G$ to be $mathbb A^1$ with a doubled origin, viewed as a group scheme over $mathbb A^1$. Then $mathcal X=[mathbb A^1/G]to mathbb A^1$ is a coarse moduli space ("there's a $B(mathbb Z/2)$ at the origin"). If we take $Dsubseteq mathbb A^1$ to be the origin, then the pullback to $mathcal X$ is the closed $B(mathbb Z/2)$ with multiplicity 1. Yet the induced morphism from $mathcal X$ to $sqrt[1]{D/mathbb A^1}cong mathbb A^1$ is not an isomorphism.

In this case, $mathcal X$ is a smooth DM stack, but has non-separated diagonal.