If $E$ and $H$ are finite-dimensional faithful representations (over $\mathbb{C}$) of a finite group $G$, with $H$ irreducible. The Molien formula describer the Poincaré series of the covariants as $$ \sum_d (\dim (\mathbb{C}[E]_d \otimes H )^G)t^d = \frac{1}{|G|}\sum_{g\in G} \frac{\operatorname{tr}(g^{-1},H)}{\det(I-tg, E)} $$
I got this reprentation of $A_5< SL(3,\mathbb{C})$ http://brauer.maths.qmul.ac.uk/Atlas/v3/matrep/A5G1-Ar3bB0 and tried to calculate (for $E=\mathbb{C}^3$) $$ \sum_d (\dim (\mathbb{C}[E]_d \otimes E)^G)t^d $$ but with the help of the formula, I got $$\dim E^G =\dim (\mathbb{C}[E]_0 \otimes E)^G= \frac{-1}{30}.$$ Every calculation was checked many times. When $H$ is the trivial representation everything works but in the case of $E$ I got nonsense...
Have you already had this issue with this formula? Do you know another way to calculate $\dim (\mathbb{C}[E]_d \otimes E)^G$ for a irreducible representation?
This formula can be found in: http://www.mccme.ru/~panyush/lecturesRT.pdf (II.3.6)
I get $\dim E^G=0$. I'll sketch my calculation so that you can see where you diverge.
For every $g$, $\det(I-tg)$ is a polynomial in $t$ with constant term $1$, and so, since for $E^G$ we're only concerned with the constant term of the Poincaré series, the formula simplifies to $$\dim(\mathbb{C}[E]_0\otimes E)^G=\frac{1}{|G|}\sum_{g\in G} \operatorname{tr}(g^{-1},E).$$
But the right hand side is just the inner product of the trivial character with the character of $E$, which is zero by orthogonality.