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Let $E$ be an elliptic curve over a field $k$ of characteristic $p = 2$ or $3$, and suppose $j(E) = 0\equiv 1728\mod p$.

In this case, $Aut(E)$ has size either 12 or 24.

Let $\omega$ be an invariant differential for $E$. How does $Aut(E)$ act on $\omega$? (Ie, it should multiply it by some nonzero element of $k$. What is that number?)

What is the structure of $Aut(E)$ as a group scheme (over $k$)?

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    $\begingroup$ By rigidity and density of the collection of finite etale $n$-torsion for $n>0$ not divisible by char($k$), the Aut-scheme for a polarized abelian variety $(A,\phi)$ over any field $k$ is finite etale. You ask about the action $G:={\rm{Aut}}_{E/k}\rightarrow k^{\times}$ on ${\rm{Lie}}(E)$. The order of $G$ is divisible by $p={\rm{char}}(k)$ and one other prime $p'$. The normal $p$-Sylow acts trivially, and the cyclic $p'$-Sylow acts faithfully by Lemma 3.3 of Edixhoven's "Neron models and tame ramification" and faithfulness of the formal group; see Wikipedia for "ss elliptic curve". $\endgroup$
    – nfdc23
    Commented Sep 25, 2016 at 0:09
  • $\begingroup$ P.s. if you are wondering what the structure of the group is over the algebraic closure as an abstract group, it is: groupprops.subwiki.org/wiki/SmallGroup(12,1) $\endgroup$
    – Maarten Derickx
    Commented Jan 14, 2021 at 17:11
  • $\begingroup$ For the benefit of Magma users who find this post in the future, I note that this group is obtained in Magma by the command SmallGroup(12,1); (M.D.'s comment leads to similar GAP code for the group, and Magma's SmallGroup identifiers are the same as GAP's) $\endgroup$
    – Michael Zieve
    Commented Sep 25, 2021 at 12:58

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There are exact formulas for the automorphisms, for example in The Arithmetic of Elliptic Curves Appendix A, proof of Proposition 1.2. From those you can write down explicitly the action of $\text{Aut}(E)$ on $\omega=dx/(2y+a_1x+a_3)$. Here's the characteristic 3 case. $E$ has a Weierstrass equation of the form $$ E:y^2=x^3+a_4x+a_6. $$ (In characteristic 3, this curve has $j(E)=0$.) Then $\text{Aut}(E)$ is the set of automorphisms of the form $$ x\mapsto u^2 x+r\quad\text{and}\quad y\mapsto u^3y $$ with $(u,r)$ satisfying $$ u^4=1\quad\text{and}\quad r^3+a_4r+(1-u^6)a_6 = 0.$$ So in this case, $\omega=dx/2y$ and the map $\phi_{u,r}$ associated to $(u,r)$ acts via $$ \phi_{u,r}^*\omega = u^{-1}\omega. $$

To answer your second question, the group scheme structure is $$ \text{Aut}(E)=\frac{k[u,r]}{\bigl(u^4-1,r^3+a_4r+(1-u^6)a_6 \bigr)}.$$

The calculation for characteristic 2, which I will leave for you, is similar.

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