Can stabilizer groups in an orbifold have global twisting?

For example, consider the two groups $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$ (where $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ is the unique nontrivial map). Both groups act on $\mathbb R$ through their common quotient $\mathbb Z$, so there are two orbifolds here: $$M_1=[\mathbb R/(\mathbb Z/3\times\mathbb Z)]$$ $$M_2=[\mathbb R/(\mathbb Z/3\rtimes\mathbb Z)]$$ The coarse spaces of $M_1$ and $M_2$ are both $S^1=\mathbb R/\mathbb Z$, and in both $M_1$ and $M_2$, all points have stabilizer group isomorphic to $\mathbb Z/3$. Are $M_1$ and $M_2$ isomorphic as orbifolds?

**Why they should be different:** Over any orbifold $X$, we think of there being space $E$ with a "nice" map $\pi:E\to X$ such that $\pi^{-1}(\{x\})$ is a $K(\Gamma_x,1)$ (see for example this paper by André Henriques). In the case of $M_1$ and $M_2$, this just means we have a $K(\mathbb Z/3,1)$ bundle over $S^1$, and its monodromy is a homomorphism $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ which should recover the difference between $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$.

This twisting is a subtle point that I don't see mentioned explicitly anywhere, and I didn't realize it could happen until now, when I actually have to do something with orbifolds and the precise definition becomes important.

End note: The answers to these related questions have very good answers in regards to defining orbifolds.