Let $X$ be a proper Deligne-Mumford stack over $\mathbb{C}$ with an action by a complex torus $T$. Let $X^T$ denote the fixed locus. 

**Question:** Is the following statement true? 

    If every point of $X^T$ is a smooth point of $X$, then $X$ is smooth.

If $X$ is a scheme, I know the answer is yes and it is proved by applying the Borel fixed-point theorem to the singular locus of $X$. So the question will be solved if the fixed-point theorem holds for DM stacks. But I could not find a reference for this.

EDIT 1: The previous version is a bit misleading: the assumption for the statement is that for any $x\in X^T$, the local ring $\mathcal{O}_{x,X}$ is regular (but not $\mathcal{O}_{x,X^T}$).

EDIT 2: Following Ariyan's comment, I present my attempted proof for the case of schemes: Let $X'$ be the singular locus of $X$. Then $T$ acts on $X'$ and leaves each of its irreducible components invariant. For each component, we apply Borel fixed-point theorem to get a fixed point $x$. By our assumption, it is a smooth point of $X$, and hence $x\not\in X'$, a contradiction. 

EDIT 3: The following result appears in Graber-Pandharipande's paper on virtual localization formula. Hope it is useful: 

If $V$ is a DM stack with a $\mathbb{C}^{\times}$-action which has no fixed points, then the equivariant Chow group $A_*^{\mathbb{C}^{\times}}(V)$ vanishes after localizing the equivariant parameter. 

Back to the original question. Suppose the singular locus $X'$ of $X$ has no fixed-points. (If it has, we win.) We apply their result to $X'$ to get the vanishing property of the equivariant Chow of $X'$. A potential contrdictation is that the fundamental class $[X']$ cannot be killed by the equivariant parameter and the reason should have something to do with the properness of $X'$. But I don't know if it is true.