Proposition 2.1.2 of Sturmfels' "Algorithms in Invariant Theory" has statement c): $\star$ is a $C[x]^\Gamma$ Module Homomorphism ie $(fI)^* = f^* I$ for $f\in C[x]$ and $I\in C[x]^\Gamma$. Where $\star$ is the Reynolds Operator $f^* = \frac{1}{|G|}\sum_{g\in G} g(f)$.
I'm not familiar with Module Homomorphisms, so I gave it a shot at proving it myself, but I get
$$(fI)^* = \frac{1}{|G|}\sum_{g\in G} g(fI)$$, but for the proof to work I would need $g(fI) = g(f)I$ which may not be the correct way to do it. In addition, I'm not sure how group actions are defined on ring multiplication. Since ring multiplication isn't a group generally, we cant seem to use any kind of group homomorphism properties. Can anyone give me some kind of general context that may help or another valid strategy?
I had another go at it with different notation, which may or may not help.
With new notation, the Reynolds Operator is:
$$f^*(\bf{x}) = \frac{1}{|G|}\sum_{g\in G}f(g.x)$$ where the group action "." acts directly on the dependence of f on $\bf{x}$. So this would make
$$(fI)(\bf{x}) = f(\bf{x})I(\bf{x})$$ so
$$(fI)^*(\bf{x}) = \frac{1}{|G|}\sum_{g\in G}f(g.\bf{x})I(g.\bf{x})$$ but since $I\in C[\bf{x}]^G$, then $I(g.\bf{x}) = I(\bf{x})$. So it can be pulled out of the sum and thus the property is proved.