The answer to the question in my title is no. An example is $\mathbb Z/5$ acting on $\mathbb C^2$ C^3$ by multiplying each coordinate by $e^{2\pi i/5}$, and acting on $\mathbb C$ by multiplication by $e^{4\pi i/5}$. In this case there are no constant or linear equivariant maps $\mathbb C^2\longrightarrow C^3\longrightarrow \mathbb C$, but any homogeneous quadratic polynomial is equivariant.
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The answer to the question in my title is no. An example is $\mathbb Z/5$ acting on $\mathbb C^2$ by multiplying each coordinate by $e^{2\pi i/5}$, and acting on $\mathbb C$ by multiplication by $e^{4\pi i/5}$. In this case there are no constant or linear equivariant maps $\mathbb C^2\longrightarrow \mathbb C$, but any homogeneous quadratic polynomial is equivariant. |
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