Suppose we have two finite vector spaces A and B. Consider a finite group G that acts on A and B via linear transformations. Assume when G acts on A and B there exists only the trivial subspaces that are invariant with respect to the action.
We want to show that if $f: A \rightarrow B$ is equivariant under the group action then either $f$ is an isomorphism or f(x) = 0, $ \forall x \in A $
Solution Attempt: They tell us only the trivial subspace can be invariant with respect to group action. Now from linear algebra we know ker($f$) is a subspace of A, and to show its invariant subspace of A we need to show ker($f$) gets mapped to an element in A. But what I am wondering is where do group actions come into play here? Because after we have show ker($f$) is invariant then we know there will only be two possibilities for it and can conclude things about f.