Let \(\mathfrak M_g\) be the stack of nodal genus \(g\) curves, and let an algebraic group \(G\) act on it trivially. Let’s understand what the points of \(\mathfrak M_g^G\) look like.
The \(S \)-points of the fixed stack \(\mathfrak M^G_g(S)\) are \(G\)-equivariant 1-morphisms \(S\to \mathfrak M_g\) where \(G \) acts on \(S\) trivially. Recall that a 1-morphism in the category of \(G\)-stacks is \((f,\sigma)\) where \(f:X\to Y\) is a \(1\)-morphism and \(\sigma\) is a \(2\)-isomorphism making the diagram
\begin{equation*}\(2\)-commute. Here \(\mu\) and \(\nu\) are simply the action maps. More concretely, if we test this against some test scheme \(T\), then for an object \((g,x)\in G(T)\times X(T)\) this \(2\)-commutativity is simply the requirement that \[\sigma_{(g,x)}:f(\mu(g, x)) \xrightarrow{\sim}\nu\big((\id_G\times f)(g,x)\big),\] or more helpfully, \[\sigma_{(g,x)}:f(g\cdot x) \xrightarrow{\sim}g\cdot f(x).\] Returning to our situation, we have a trivial action on our test scheme \(S\) and our stack \(\mathfrak M_g\), so both \(\mu\) and \(\nu\) are trivial for us. Testing against a \(T\) and an object \((g,x)\in G(T)\times S(T)\) again, a \(G\)-equivariant morphism \((f,\sigma)\) from \(S\) to \(\mathfrak M_g\) is one satisfying \[\sigma_{(g,x)}:f(g\cdot x)\xrightarrow{\sim}g\cdot f(x)\] as we had above, but here the group action is trivial, so really we’re asking that \[\sigma_{(g,x)}:f(x)\xrightarrow{\sim}f(x).\] This is to say, \(\sigma_{(g,x)}\in \Aut(f(x))\).
We have further compatibility requirements of course dictating the various ways we can resolve expressions like \(g_1\cdot g_2\cdot ... \cdot g_n \cdot x\), which in the case of \(\sigma\) reduces to requiring that \(\sigma_{(g,x)}\circ\sigma_{(h,x)} = \sigma_{(gh,x)}. \) This means \(\sigma_{(-,x)}\) is a group homomorphism from \(G\) to \(\Aut(f(x))\). In any case, a morphism \(S\to \mathfrak M_g\) is the same data as a family of genus \(g\) nodal curves \(E\to S\), and an automorphism of the point \(f(x)\) in \(\mathfrak M_g\) corresponds to an automorphism of the fiber curve \(E_x\) over \(x\in S\). This leads us to the conclusion that objects in the fixed stack \(\mathfrak M_g^G\) are exactly families of nodal curves with fiberwise \(G\)-actions.
In particular, this means that the stack \(\mathfrak M^G_g\) is bigger than the stack \(\mathfrak M_g\), in some sense, because if there is ever a nontrivial fiberwise action we can place on a family \(E\to S\) then we get two objects in \(\mathfrak M^G_g\) both mapping to the same object in \(\mathfrak M_g\) under the forgetful map.
As a corollary of the stuff discussed in this example we get the following.
Corollary
Let \(X\) be a DM stack locally of finite type over a field \(k\) and let \(T\) be an algebraic torus acting on \(X\). Then the natural map \(\iota:X^{hT}\to X\) is a closed embedding.
Proof
Proposition A.12 from Aranha et. al.’s “Virtual Localization Revisited” says that the inclusion \(X^T\hookrightarrow X \) of the fixed substack \(X^T \) is a closed immersion when \(X\) is a quasi-DM stack locally of finite type over \(k\) and \(T\) has connected fibers over the base. The natural map \(\iota:X^{hT}\to X\) factors through this inclusion, and the map \(X^{hT}\to X^T\) is surjective and fails to be an isomorphism if and only if we have too many maps \(T\to \Aut(x)\) for points \(x\in X\). As we saw above, such morphisms are the reason \(\iota\) fails to be a monomorphism in general. However, when \(X\) is Deligne-Mumford, the stabilizer groups of points in \(X\) are all finite, and hence the only morphism \(T\to \Aut(x)\) for any point \(x\in X\) is the trivial map.