Fundamental group of $M_g^\circ$ Let's work over the complex numbers $\mathbb{C}$. Let $g\geq3$ be an integer. Let $\mathcal{M}_g$ be moduli stack of smooth genus $g$ curves. Let $M_g$ be the corresponding coarse moduli scheme. They share an open subscheme $M_g^\circ$ parametrizing  automorphism-free smooth genus $g$ curves. 
What is the fundamental group $\pi_1(M_g^\circ)$?
How does it compare to the following notions:
(1) The orbifold mapping class group $\pi_1(\mathcal{M}_g)$, 
(2) The topological fundamental group  $\pi_1(M_g)$, 
(3) The mapping class group of a genus $g$ surface, $\mathrm{Mod}_g$?
 A: Since $\mathcal{M}_g=[\mathcal{T}_g/\text{Mod}_g]$, and $\mathcal{T}_g$ Teichmüller space is contractible, (1) and (3) are going to be isomorphic.
This was mentioned previously on MathOverflow here, or can be found for instance on page 361 of Farb and Margalit "A primer on Mapping Class Groups".
Meanwhile, (2), the fundamental group of the coarse moduli space $\pi_1(M_g)$, is trivial, as described by Andy Putman's answer to a question here.
So, we really just need to compare $\pi_1(M^\circ_g)$ to $\pi_1(\mathcal{M}_g)$.  The inclusion map will give us a homomorphism from $\pi_1(M^\circ_g)\to \pi_1(\mathcal{M}_g)$, and presumably we can say something nice about this but I don't know anything off the top of my head and have run out of time to research it...
A: Except in a few trivial cases, the locus of curves which have an extra automorphism will have codimension greater than one in $\mathcal M_g$. When that happens, the fundamental group of $\mathcal M_g$ must equal the fundamental group of $\mathcal M_g^{\circ}$. (To see this, one can pass to the $3$-torsion cover of $\mathcal M_g$, which is a smooth scheme, and use the fact that removing a codimension two subscheme of a smooth scheme does not affect $\pi_1$.) So it equals cases 1 and 3.
