I believe QiL in that the result is well-known. The point of this answer is to show, given the data in the OP, that the automorphism group is $SL_2(\mathbb{F}_3)$ from the first principles. I enjoyed this exercise as a refresher, so...
The point doubling formula for this $j=0$ curve is the following. If $P=(x,y)$ is an affine point, then $2P=(x^4,x^3+x^6+y+1)$. Similarly the additive inverse is given by the formula $-P=(x,y+1)$. These formulas are given in all books on elliptic curves. An affine point is thus 3-torsion, iff $2P=-P$, iff $x=x^4$ and $y=x^3+x^6+y$. The first equation simply means that $x\in\mathbb{F}_4$. But then we have either $x^3=0$ or $x^3=1$, and the latter equation thus holds automatically. Therefore the $\mathbb{F}_4$-rational affine points $(0,0); (0,1)$; $(1,\zeta);(1,\zeta^2)$; $(\zeta,\zeta);(\zeta,\zeta^2)$;$(\zeta^2,\zeta);(\zeta^2,\zeta^2)$ are exactly the 3-torsion points.
The sum of the points $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ is $P+Q=(x_3,y_3)$, where
$\lambda=(y_1+y_2)/(x_1+x_2)$, $x_3=\lambda^2+x_1+x_2$, $y_3=\lambda(x_1+x_3)+y_1+1$. So if we write $P_1=(0,0)$, and $P_2=(1,\zeta)$, then a calculation repeatedly using the relation $\zeta^2=\zeta+1$ shows (barring my mistake) that $P_3=P_1+P_2=(\zeta,\zeta)$ and $P_4=P_1+2P_2=(\zeta^2,\zeta^2)$. Thus we see that $P_1$ and $P_2$ generate all of $E[3]$. The other 3-torsion points are the negatives of the listed ones.
Any automorphism of $E$ will act on the 3-torsion $E[3]$, so let us
consider the action of the listed automorphisms on $E[3]\simeq\mathbb{F}_3^2$ (isomorphic as abelian groups). I write them as 2x2 matrices with entries in $\mathbb{F}_3$ with respect to the basis $\{P_1,P_2\}$. The automorphism $b$ maps all the points to their negatives, and thus maps to $-I_2$. The automorphism $a$ maps $P_1=(0,0)$ to itself, and $P_2=(1,\zeta)$ to $(\zeta,\zeta)=P_1+P_2$. The matrix is thus
$$
a\mapsto \pmatrix{1&1\cr0&1\cr}.
$$
Continuing this we see that $c_1(P_1)=c_1(0,0)=(0+1,0+0+\zeta)=(1,\zeta)=P_2$ and
$c_1(P_2)=c_1(1,\zeta)=(1+1,\zeta+1+\zeta)=(0,1)=-P_1$. Therefore
$$
c_1\mapsto \pmatrix{0&-1\cr 1&0\cr}.
$$
In the same way we see that $c_\zeta(P_1)=P_1+P_2$ and $c_\zeta(P_2)=P_1+2P_2$, so
$$
c_\zeta\mapsto \pmatrix{1&1\cr1&2\cr}.
$$
As a further check we can compute that the relation $ac_1a^{-1}=c_\zeta$ (see my comments to OP) holds on the matrix side as well.
It is then easy to check that the listed matrices generate all of $SL_2(\mathbb{F}_3)$.
Because we knew that the automorphism group $G$ is of order $24=|SL_2(\mathbb{F}_3)|$, we
have seen that $G\simeq SL_2(\mathbb{F}_3)$ and that elements of $G$ can be identified by their action on $E[3]$ (IOW the above homomorphism sending an automorphism of $E$ to its restriction on $E[3]$ is injective).
I think that it always happens that the restriction of the automorphism group on $E[p]$ for any prime $p$ is inside the special linear group. This is because the action has to respect the Weil pairing.
