It is known that the linear transformation fixing $f_0$ are exactly the transformations $(x_1,x_2,x_3)\mapsto (\lambda x_{\sigma(1)}, \mu x_{\sigma(2)}, (\lambda\mu)^{-1}x_{\sigma(3)})$, where $\sigma\in\mathfrak{S}_3$ and $\lambda^3=\mu^3=1$.
These transformations also satisfy $f_i^\sigma=\gamma_if_i$ for some $\gamma_i$ such that $\gamma_i^3=1$, for $i=1,2$. Hence $G$ consists exactly of these $54$ transformations. Notice they are all unitary.
Now for such a transformation to be in $H$, you need $\lambda^2\mu=1$ and $\lambda\mu^2=1$, that is $\lambda=\mu$, so the elements of $H$ are the transformations $(x_1,x_2,x_3)\mapsto (\lambda x_{\sigma(1)}, \lambda x_{\sigma(2)}, \lambda x_{\sigma(3)})$, where $\sigma\in\mathfrak{S}_3$ and $\lambda^3=1$, which has $18$ elements, so $H$ is a proper subgroup of $G.$ (Note that it is easy to understand the group structure of $G$ and $H$, but I'm a bit lazy now).
Anyway, a polynomial fixed by $G$ or $H$ will have to be symmetric, since $G$ and $H$ contain the symmetric group on three letters, so you cannot expect $f_1$ or $f_2$ to be fixed by $H$. If I have time, I'll do the computations later today.