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 $k$)?



Automorphisms of elliptic curves in characteristic 2 and 3
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.


