For $\sigma \in \mathrm{GL}\_n(\mathbb C)$ and $f(x\_1,...,x\_n)\in \mathbb C[x\_1,...,x\_n]$, let $f^ \sigma (x):=f(\sigma^{-1}x)$, for $x=(x\_1,...,x\_n)$. 


For a subgroup $G$ of $\mathrm{GL}\_n(\mathbb C)$, let $\mathbb C[x\_1,...,x\_n]^G :=\{f\in \mathbb C[x\_1,...,x\_n] : f^\sigma =f ,\forall \sigma \in G\}$.


Now let $f\_0,f\_1,f\_2 \in \mathbb C[x,y,z]$ be defined as $f\_0(x,y,z)=x^3+y^3+z^3+6xyz$, $$f\_1(x,y,z)=3(x^2y+y^2z+z^2x),\quad f\_2(x,y,z)=3(xy^2+yz^2+zx^2).$$ 


Let $G=\{\sigma \in \mathrm{GL}\_3(\mathbb C) : f\_0^\sigma =f\_0,\; (f\_1^\sigma)^3=f\_1^3,\; (f\_2^\sigma)^3=(f\_2)^3 \}$ .


Let $H=\{\sigma \in \mathrm{GL}\_3(\mathbb C) : f\_0^\sigma =f\_0,\; f\_1^\sigma=f\_1,\; f\_2^\sigma=f\_2 \}$.


What are the orders of $G$ and $H$ ? Is $H$ a proper subgroup of $G$ ? How to show that every matrix in $G$ is unitary ? Is it true that $\mathbb C[x,y,z]^G=\mathbb C[f\_0,f\_1^3,f\_2^3]$ ? Is it true that $\mathbb C[x,y,z,]^H=\mathbb C[f\_0,f\_1,f\_2]$ ?



Ring of invariants of some special type of subgroups of $GL_3(\mathbb C)$
**Note added on 26 Nov 2018:** I have corrected my answer, which had a serious mistake.


For simplicity of notation, let $(x,y,z) = (x\_1,x\_2,x\_3)$. The Hessian form associated to $f\_0 = {x\_1}^3+{x\_2}^3+{x\_3}^3+6x\_1x\_2x\_3$ is 
$$
H(f\_0) = \frac{\partial^2f\_0}{\partial x\_i\partial x\_j}\,\mathrm{d}x\_i\circ\mathrm{d}x\_j\,.
$$
The determinant of this Hessian form is easily computed to be
$$
\Delta = -6^3 ({x\_1}^3+{x\_2}^3+{x\_3}^3-3x\_1x\_2x\_3)\,
(\mathrm{d}x\_1\wedge\mathrm{d}x\_2\wedge\mathrm{d}x\_3)^{\otimes2}.
$$
Any linear transformation of $\mathbb{C}^3$ that preserves $f\_0$ must preserve $\Delta$. Since 
$$
{x\_1}^3{+}{x\_2}^3{+}{x\_3}^3{-}3x\_1x\_2x\_3 = (x\_1{+}x\_2{+}x\_3)(x\_1{+}sx\_2{+}s^2x\_3)(x\_1{+}s^2x\_2{+}sx\_3)
$$
where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f\_0$ must must permute and scale the three linearly independent forms
$$
\xi\_i = (x\_1{+}s^ix\_2{+}s^{2i}x\_3)\quad i = 0,1,2.
$$
Thus, $L^\*(\xi\_i) = \lambda\_i\,\xi\_{\pi(i)}$ for $i=0,1,2$, where the $\lambda\_i$ are nonzero and $\pi$ is a permutation of $\{0,1,2\}$. Expressing $f\_0$ in terms of the $\xi\_i$, we find that $f\_0 = \tfrac13({\xi\_0}^3{+}{\xi\_1}^3{+}{\xi\_2}^3)$.
It now follows that $\lambda\_i^3 = 1$ for $i=0,1,2$. 


Conversely, for any permuation $\pi$ of $\{0,1,2\}$ and any cube roots of unity $\lambda\_i$ for $i=0,1,2$, the formula $L^\*(\xi\_i) = \lambda\_i\,\xi\_{\pi(i)}$ defines a linear transformation of $\mathbb{C}^3$ that preserves $f\_0$.
Thus, the stabilizer of $f\_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $F=|\xi\_1|^2{+}|\xi\_2|^2{+}|\xi\_3|^2 = 3(|x\_1|^2{+}|x\_2|^2{+}|x\_3|^2)$).
Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).


Meanwhile, we find
$$
f\_1 = \tfrac13({\xi\_0}^3+s^2\,{\xi\_1}^3+s^4\,{\xi\_2}^3)
\qquad\text{and}\qquad
f\_2 = \tfrac13({\xi\_0}^3+s\,{\xi\_1}^3+s^2\,{\xi\_2}^3).
$$
It follows that $L$, as defined above, preserves $(f\_i)^3$ if and only if $\pi$
is an *even* permutation of $\{0,1,2\}$. Thus, $G$ has order $81$.


In order to preserve $f\_1$ and $f\_2$, the permutation $\pi$ must be the identity.
Thus, $H$ has order $27$.


One sees that $\mathbb{C}[x,y,z]^G$ contains $f\_1f\_2$, so it is not equal to $\mathbb{C}[f\_0,(f\_1)^3,(f\_2)^3]$.


On the other hand $H$ is the group that preserves each of the ${\xi\_i}^3$ (since these are linear combinations of $f\_0,f\_1,f\_2$ and *vice versa*)
and $H$ preserves a monomial ${\xi\_0}^{n\_0}{\xi\_1}^{n\_1}{\xi\_2}^{n\_2}$ if and only if all of the $n\_i$ are divisible by $3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[{\xi\_0}^3,{\xi\_1}^3,{\xi\_2}^3]=\mathbb{C}[f\_0,f\_1,f\_2]$.


I think that answers all your questions.


