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I am looking the group of automorphisms $G$ of the curve defined over $\mathbb F_3$ by (in projective coordinates) $Y^2Z=X(X-Z)(X-2Z)$. Obviously, there are the automorphisms $X\mapsto X+\alpha Z$ (for $\alpha\in\mathbb F_3$) and $Y\mapsto\pm Y$. But are they the only ones? And a second question: What is the field $(\mathbb F_3(x)[y])^G$? I did not manage to determine it.

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This is the elliptic curve $E:y^2=x^3-x$ over $\mathbb F_3$ whose $j$-invariant is $0$. The automorphism groups of elliptic curves over finite fields are quite well known, so this question is probably better suited for MathStackExchange. Over $\overline{\mathbb F}_3$ the automorphism group has order 12; it is the semi-direct product of $C_4$ and $C_3$. The "other" automorphism, i.e., the one you need to compose with the ones that you already have, is $(x,y)\mapsto(-x,iy)$, where $i^2=-1$. but since $-1$ is not a quadratic residue mod 3, you need to go to the field $\mathbb{F}_9$ to get all of $\operatorname{Aut}(E/ \overline{\mathbb F}_3)$.

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    $\begingroup$ The OP asked for automorphisms of the curve, not of the elliptic curve -- i.e. the origin may move, so the full group of curve automorphisms is the semidirect product of $E$ with its group of elliptic-curve automorphisms. $\endgroup$ Commented Apr 24, 2019 at 3:18
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A generator of the field of $G$-invariant rational functions on the curve is given by \begin{equation*} f = \frac{(x^3-x)^3}{(x^2+1)^2 (x^2+x-1)^2 (x^2-x-1)^2} \end{equation*} Indeed, since $G$ contains the elliptic involution $(x,y) \to (x,-y)$, this field is contained in $\mathbb{F}_3(x)$. The group $E(\mathbb{F}_3)$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$. Let $H$ be the group generated by the transformations $x \to x+a$ with $a \in \mathbb{F}_3$, and by those induced by the $\mathbb{F}_3$-rational translations on $E$. The degree of the extension $\mathbb{F}_3(x)/\mathbb{F}_3(x)^H$ is equal to $|H|=12$.

Note that $H$ acts on $\mathbb{P}^1(\mathbb{F}_9)$ with two orbits: $\mathbb{P}^1(\mathbb{F}_3)$ and its complement. The respective stabilizers have order 3 and 2 respectively. Let $f$ be a generator of the field $\mathbb{F}_3(x)^H$; it is a rational function of degree 12. Let $a$ (resp. $b$) be the value of $f$ on $\mathbb{P}^1(\mathbb{F}_3)$ (resp. its complement). The element $b \in \mathbb{P}^1(\mathbb{F}_9)$ is invariant under $\mathrm{Gal}(\mathbb{F}_9/\mathbb{F}_3)$ and must be distinct from $a$ since $f$ has only degree $12$. Composing $f$ with an homography, we may assume $a=0$ and $b=\infty$, which gives the function $f$ above.

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