In defining $\overline{M_{g,n}}$, we say that an $S$point of it is a nodal curve $X \to S$, and $n$ sections $s_i : S \to X$, $i = 1,\cdots, n$ and plus stability conditions. Implicitly, we are ordering these sections. What if we don't order them and take the following as definition. An $S$point is a nodal curve $X \to S$ together with an effective relative Cartier divisor $D$ on $X$ of degree $n$, plus the same stability condition on each geometric fiber. What kind of moduli space shall we get with this definition? Is it just the usual $\overline{M_{g, n}}$ quotient by the order $n$ symmetric groups?

I believe the answer is yes, both if you think of stacks and coarse spaces. I don't understand Jason Starr's comment, but maybe I am missing something obvious. To prove it on the level of stacks, the biggest difficulty is maybe sorting out the formalism. Let me try to spell out some details here. Consider the functor $F$ which maps a scheme $S$ to the groupoid whose objects are $n$pointed stable curves $(X,\sigma_1,...,\sigma_n)$ over $S$ and whose morphisms are pairs $(f,\pi)$ with $f : X \to X'$ an isomorphism over $S$ and $\pi \in \mathbb S_n$ a permutation, such that $f \circ \sigma_i = \sigma_{\pi(i)}'$ for all $i$. In other words $F(S)$ is the action groupoid of $\mathbb S_n$ acting on $\overline M_{g,n}(S)$. Then the stackification of $F$ is $[\overline M_{g,n}/\mathbb S_n]$. Let $G$ be the functor you propose, $G(S)$ is the groupoid whose objects are pairs $(X,D)$ of a curve over $S$ and a relative effective multiplicity free Cartier divisor of degree $n$, satisfying the obvious stability condition. Morphisms are isomorphisms $f \colon X \to X'$ over $S$ with $f(D)=D'$. There is a natural transformation $F \to G$ given by letting $D$ be the union of the schemetheoretic images of the sections $\sigma_i$. Since $G$ is a stack we get an induced map from the stackification of $F$ to $G$. The crucial point in showing that this is an isomorphism is the following: pick $(X,D)$ in $G(S)$. Since $D \to S$ is finite étale, there is an open cover $S' \to S$ such that $D \times_S S' \to S'$ is trivial. Then the pullback of $(X,D)$ to $S'$ is in the essential image (in fact in the image) of $F(S') \to G(S')$. 

