TITLE: What is $Aut(Ell)$?
QUESTION [19 upvotes]: Consider the stack $Ell$ (of groupoids) of elliptic curves. I'm interested in the autoequivalence 2-group of $Ell$, the objects of which consists of transformations $Ell \Rightarrow Ell: Ring \to Gpd$ valued in equivalences of groupoids. The arrows are isomorphisms of such transformations.
In a chat opinions ranged from optimistic that it would be large to hunches it would be small (in fact $\mathbb{Z}/2 \rightrightarrows \ast$)
Even if this 2-group is very small, I would also be interested in knowing if it's possible to calculate much of the endomorphism monoidal groupoid of $Ell$, given that we know a very explicit presentation of it (using a Hopf algebroid built from finitely generated polynomial rings). Here we would take all transformations, not just those valued in equivalences.
EDIT, Will Jagy: to get fuller context, you can scroll arbitrarily, and conveniently, back in the transcript http://chat.stackexchange.com/transcript/9417/2013/6/28/5-16 where David's announcement of this question occurs pretty late in this segment. I will check later today, the terminal hour marker '16' may change, that is how the system works.

REPLY [3 votes]: The 2-group is $B\mathbb Z/2$. In other words, the automorphism $1$-group of $M_{1,1}$ is trivial, and the identity functor $M_{1,1} \to M_{1,1}$ has exactly one non-identity invertible natural transformation to itself: the one which sends a family of elliptic curves $\xi \colon E \to S$ to $\xi \circ i \colon E \to S$, where $i$ is inversion in the group structure of $E$.
The claim about the $1$-group was explained in the comments: the automorphism $1$-group of $\overline M_{1,1}$ is $\mathbb G_m$, since $\overline M_{1,1} \cong \mathbb P(4,6)$. None of these automorphisms fix the point at infinity. So we need only to determine the natural equivalences from the identity to itself. 
Such a natural equivalence would assign to any $\newcommand{\id}{\mathrm{id}}\xi \colon E \to S$ an automorphism $a_\xi \colon E \to E$ over $S$, and it should satisfy the conditions of a natural transformation. For any $E \to S$ the automorphism group of $E$ over $S$ has two distinguished elements, $\id$ and $-\id$, and these are stable under pullback. In particular given $S' \to S$ and $\xi' \colon E' \to S'$ the pullback of $\xi$, if $a_\xi$ is trivial or inversion in the group, then the same holds for $a_{\xi'}$. The converse holds by descent if $S' \to S$ is étale. 
Now let $X \to M_{1,1}$ be an étale cover by a scheme, let $\eta \colon C \to X$ be the pullback of the universal family. There is an open dense $U \subset X$ such that the only automorphism of $C$ over $U$ is inversion in the group. Then the same holds globally on $X$, so $a_\eta = \pm \id$, because the isomorphism scheme of $C$ over $X$ is separated. Let $\xi \colon E \to S$ be arbitrary. There is an étale cover $S' \to S$ such that $\xi'$ is pulled back from $\eta$. Then $a_{\xi'}$ is $\pm \id$. Then the same is true for $a_\xi$.