How to show a subspace is invariant with respect to a group action 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. 
 A: The reason you can conclude things about $f$ is from studying the group action. The very statement "the kernel of $f$ is invariant" is a statement about the group action. Furthermore, to show this statement is true, one must use the fact that $f$ is $G$-equivariant. Without it, all hope is lost.
As anon mentions, $f(x) = cx$ doesn't (usually) make sense. I believe you meant that $f$ is an isomorphism, and possibly thinking about the case $A = B = \mathbb{R}$.
Edit:
To show the claim, that $\ker f$ is invariant under the $G$-action, suppose that  $\alpha \in \ker f$. Then we want to show that $g \cdot \alpha$ is also in the $\ker f$. This follows immediately from $f$ being $G$-equivariant by writing $f(g \cdot \alpha) = g \cdot f(\alpha)$.
A: Here's a counterexample. Let $A=B=\mathbb C$ and $G=\{1,i,-1,-i\}$, so that $G$ acts on $A$ and $B$ by multiplication. Define $f:A\to B$ by $f(z)=iz$. Then $f$ is equivariant under the actions of $G$ on $A$ and $B$, but $f$ is not trivial. Moreover, if we think of $\mathbb C$ as a 2-dimensional real vector space, $f$ is not multiplication by a scalar.
