How the automorphism group of an elliptic curve acts at the localization of the stack $\mathcal{M}_{1, 1, k}$ at the corresponding point I am studying the enlightening article "The Picard Group of $\mathcal{M}_{1, 1, S}$", written by Fulton and Olsson, but I have some problems with a proof.
Setting
Let $\mathcal{M}_{1, 1, k}$ denote the stack of elliptic curves over an algebraically closed field $k$ with $char(k)=3$. We have
\begin{equation*}
V = Spec(k[\lambda][1/(\lambda(\lambda - 1))]) \rightarrow \mathcal{M}_{1, 1, k} \rightarrow \mathbb{A}^{1}_{k}
\end{equation*}
where the first map (which is étale) is the Legendre family, hence induced by the ellptic curve $E_{V} \rightarrow V$
\begin{equation*}
E_{V} \colon Y^{2}Z = X(X-Z)(X - \lambda Z)
\end{equation*}
and the second map is the morphism into the coarse moduli space induced by the $j$-invariant. The composition is induced by the ring map sending $j \mapsto \frac{2^{8}(\lambda^{2}-\lambda+1)^{3}}{\lambda^{2} (1 - \lambda)^{2}}$.
Recall that in $char(k)=3$ the elliptic curves have automorphism group $\mathbb{Z}/2\mathbb{Z}$ except that for the curve with $j$-invariant equal to $0$ (which in the family corresponds to $\lambda = -1$) in which case the automorphism group is isomorphic to $G = \mathbb{Z}/4\mathbb{Z} \ltimes \mathbb{Z}/3\mathbb{Z}$.
The problem
I want to understand how the group $G$ acts on the localization in $\mathcal{M}_{1, 1, k}$ at the point corresponding to the curve whith automorphism group $G$. Since $V \rightarrow \mathcal{M}_{1, 1, k}$ is étale, the problem is equivalent to understanding how it acts on the localization at $\lambda = -1$ in V.
What it is said in the article
For comfort, we can translate and consider $\mu = \lambda + 1$ so the localization is at $\mu = 0$ and we have the morphism $Spec(k[[\mu]]) \rightarrow Spec(k[[j]])$ induced by $j \mapsto \frac{\mu^{6}}{(\mu^{4}-1)}$ (since we hare in $char(k)=3$).
Now the group $G$ acts on $k[[\mu]]$. The group sits in an exact sequence
\begin{equation*}
1 \rightarrow \{ \pm 1 \} \rightarrow G \rightarrow S_{3} \rightarrow 1
\end{equation*}
and the action on $k[[\mu]]$ factors through the action of $S_{3}$ on $k[[\mu]]$ given by the two automorphisms
\begin{equation*}
\alpha \colon \mu \mapsto -\mu
\end{equation*}
and
\begin{equation*}
\beta \colon \mu \mapsto \mu/(1-\mu) = \mu(1 + \mu + \mu^{2} + \ldots).
\end{equation*}
My question is: How the authors of the article determine the actions of $\alpha$ and $\beta$?
My attempt at a solution
I think to know how $\alpha$ acts in such a way.
First of all, if we consider a model for an elliptic curve with $j$ invariant equal to 0, like $y^{2}=x^{3} + a_{4}x + a_{6}$, with $a_{4} \ne 0$, the morphism of elliptic curves are of the form
\begin{equation*}
\begin{split}
x &\mapsto u^{2}x' + r \\
y &\mapsto u^{3}y'
\end{split}
\end{equation*}
and imposing the equality between the curve in $(x, y)$ and the curve in $(x', y')$, i.e. imposing the morphism being an automorphism, we obtain $u^{4}=1$ and $r^{3} + a_{4}r + (1-u^{2})a_{6}=0$. The $12$ elements $(u, r)$ which satisfy the equations are the group elements of $G$. In particular $\alpha = (i, 0)$.
To see how $\alpha$ acts:
let $Spec(k[x, y])/(y^{2} = x^{3} + a_{2}x^{2} + a_{4}x + a_{6})) \rightarrow Spec(k)$ be a generic elliptic curve. On the coefficients $(a_{2}, a_{4}, a_{6})$, the transformation $(u, r)=(i, 0)$ acts via $(i, 0) \star (a_{2}, a_{4}, a_{6}) = (i^{2}a_{2}, i^{4}a_{4}, i^{6}a_{6})$. Since we are interested at the point $(0, 1, 1)$ we consider the curve $y^{2} = x^{3} + \mu x^{2} + x + 1$ on which the action is $(i, 0) \star (\mu) = i^{2} \mu = - \mu$ and taking the localization at $\mu=0$ is exactly what we want to see the action of $\alpha= (i, 0)$ at the point $\lambda = -1$ (i.e. $\mu=0$).
My problem is that the previous argument cannot be applied to $\beta$.
Can somone help me? (maybe also with a simpler way to compute the action of $\alpha$)
Thanks
 A: From Section 3 of the paper, we see that the moduli stack of elliptic curves (over a ring $A$) has a quotient stack presentation $$ \mathcal{M}_{1,1,A} \simeq [U/G] $$ where $U = \operatorname{Spec} A[a_{1},a_{3},a_{2},a_{4},a_{6},\Delta^{-1}]$ and $G = \operatorname{Spec} A[u^{\pm},r,s,t]$. The group law on $G$ and the action of $G$ on $U$ come from making the change-of-variable \begin{equation} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \mapsto \begin{bmatrix} u^{2} & & r \\ u^{2}s & u^{3} & t \\ & & 1 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix} \tag{1}\label{eqn01} \end{equation} in the Weierstrass equation $Y^{2}Z+a_{1}XYZ+a_{3}YZ^{2} = X^{3} + a_{2}X^{2}Z + a_{4}XZ^{2} + a_{6}Z^{3}$ (see Silverman, Arithmetic of Elliptic Curves, Section III.1, Table 3.1).
Suppose $2 \in A^{\times}$. Given $a_{1},a_{3},a_{2},a_{4},a_{6} \in A$ such that $\Delta \in A^{\times}$, we can change variables so that $a_{1} = a_{3} = 0$ (which implies $s = t = 0$), and furthermore we can replace $A$ by an etale cover $A \to A'$ and change variables so that \begin{equation} \begin{aligned} a_{6} &= 0 \\ a_{2}+a_{4}+1 &= 0 \end{aligned} \tag{2}\label{eqn02} \end{equation} (i.e. write the equation in (dehomogenized) Legendre normal form \begin{equation} y^{2} = x(x-1)(x-\lambda) \tag{3}\label{eqn03} \end{equation} for $\lambda = a_{4}$). The formula \eqref{eqn01} becomes $(x,y) \mapsto (u^{2}x+r,u^{3}y)$. The equation \eqref{eqn03} is transformed into \begin{equation} \textstyle y^{2} = (x+\frac{r}{u^{2}})(x+\frac{r-1}{u^{2}})(x+\frac{r-\lambda}{u^{2}}) \tag{4} \label{eqn04} \end{equation} and imposing the condition \eqref{eqn02} gives \begin{equation} \begin{aligned} r(r-1)(r-\lambda) &= 0 \\ \textstyle (1+\frac{r}{u^{2}})(1+\frac{r-1}{u^{2}})(1+\frac{r-\lambda}{u^{2}}) &= 0 \end{aligned} \tag{5} \label{eqn05} \end{equation} respectively.
Let $k$ be an algebraically closed field of $\operatorname{char} k = 3$, and set $A := k[[\mu]]$ and $\lambda = \mu-1$. The stabilizer $\Gamma_{\overline{x}}$ can be identified with the set of pairs $(u,r) \in A^{\times} \times A$ satisfying \eqref{eqn05}, and the action $\Gamma_{\overline{x}} \times A \to A$ is defined by \begin{equation} (u,r) \cdot \lambda := \textstyle \frac{1}{u^{4}}(r(r-1) + r(r-\lambda) + (r-1)(r-\lambda)) \end{equation} (namely the coefficient of $x$ in \eqref{eqn04}, the new $a_{4}$). You can check that the solutions are \begin{equation} (u,r) = (\pm 1 , 0), (\pm \sqrt{\lambda},0), (\pm \sqrt{-1},1), (\pm \sqrt{\lambda-1},1), (\pm \sqrt{-\lambda},\lambda),(\pm \sqrt{1-\lambda},\lambda) \end{equation} which sends $\mu$ to \begin{equation} \mu,\frac{\mu}{\mu-1},-\mu,\frac{\mu}{\mu+1},\frac{-\mu}{\mu-1},\frac{-\mu}{\mu+1} \end{equation} respectively. It seems the authors don't specify what $\alpha,\beta$ are but if they're interested in checking a certain condition for a set of generators of $S_{3}$, then it is enough to choose elements of order 2 and 3.
