Automorphisms of an elliptic curve fixing the invariant differential? If we consider an elliptic curve $E/k$ given in Weierstrass form $y^{2}+a_{1}xy+a_{3}y=x^{3}+a_{2}x^{2}+a_{4}x+a_{6}$, then I know that the translation maps $\tau_{P}$ with $P\in{E}$ fix the invariant differential $\omega_{E}=\frac{dx}{2y+a_{1}x+a_{3}}$, i.e. $\tau_{P}^{*}\omega_{E}=\omega_{E}$. But I'm wondering if the converse is true? Namely, if $\phi:E\rightarrow{E}$ is an automorphism of $E/k$ such that $\phi^{*}\omega_{E}=\omega_{E}$, then $\phi$ should be a translation map?
My only attempt is assuming that $\phi$ is of the form $(x,y)\mapsto{(u^{2}x+r,u^{3}y+u^{2}sx+t)}$ for some $u\in{k^{*}}$ and  $r,s,t\in{k}$ (I think we can assume this) and applying the definition directly
$\phi^{*}\frac{dx}{2y+a_{1}+a_{3}}=\frac{d\phi^{*}x}{2\phi^{*}y+a_{1}\phi^{*}x+a_{3}}=\frac{d(u^{2}x+r)}{2(u^{3}y+u^{2}sx+t)+a_{1}(u^{2}x+r)+a_{3}}=\frac{u^{2}dx}{2u^{3}y+(2u^{2}s+a_{1}u^{2})x+(2t+a_{1}r+a_{3})}$
which by hypothesis is equal to $\omega_{E}$ again. This seems to suggest that $u=1$ and $s=0$, and also $2t+a_{1}r=0$. So $\phi$ is of the form $(x,y)\mapsto{(x+r,y+t)}$. Is it possible to conclude from this that $\phi$ is a translation map? How can I do that?
 A: [Corrected (mistake in characteristic 2: multiplication by $-1$
always fixes $\omega_E$)]
The converse is true, except in characteristic 2 and for 
some supersingular $E/k$ in characteristic 3.
Composing $\phi$ with translation by $-\phi(0)$ yields an automorphism
$\psi: E \to E$ sending $0$ to $0$ and inducing the same action on $\omega_E$
as the $\phi$.  Let $A$ be the group of automoprhisms sending $0$ to $0$.
Assume first that $E$ is not a supersingular curve
in characteristic $2$ or $3$.  Then $A$ is cyclic,
usually of order $2$ except that curves with $j=1728$ and $j=0$ have
an automorphism of order $4$ or $6$ if $k$ contains the correpsonding
roots of unity.  In this case the homomorphism $A \to k^*$ obtained from
the action on $\omega_E$ is injective, except in characteristic 2
where $-1=1$ so the multiplication-by-$(-1)$ map also fixed $\omega_E$.
If $k$ is algebraically closed of characteristic $p=2$ or $3$,
and $E$ is supersingular, then $A$ is non-abelian of order 
$24$ or $12$ respectively, so the map to $k^*$ cannot be injective 
because a finite subgroup of $k^*$ is cyclic.  For $p=2$ the kernel
still contains $-1$ as before, but also six further automorphisms,
for a total of $8$, which constitute a quaternion group; the image of
$A$ in $k^*$ consists of the cube roots of unity.  For $p=3$ the kernel 
is cyclic of order $3$, and the image consists of the 4th roots of unity;
but if $k$ is not algebraically closed then the kernel might still be trivial.
Explicit examples of non-identity $\psi$ with $\psi^* \omega_E = \omega_E$:
In characteristic 2, take $E: y^2 + y = x^3$, with $\omega_E = dx$.
Let $\psi(x,y) = (x+1, y+x+\eta)$ where $\eta^2 + \eta = 1$.
Note that $\psi^2$: $(x,y) \mapsto (x,y+1)$is multiplication by $-1$.
In characteristic 3, take $E: y^2 = x^3 - x$, with $\omega_E = dx/y = dy$.
Let $\psi(x,y) = (x+1, y)$.  However, for $y^2 = x^3 + x$ over 
the prime field ${\bf Z} / 3{\bf Z}$, the kernel is trivial
because the only nontrivial automorphisms fixing both $0$ and $\omega_E$
are $(x,y) \mapsto (x \pm i,y)$ with $i^2 = -1$.
