$ \operatorname{Ker}TR = \{0\}$ but $ \operatorname{Ker}T \not= \{0\}$ where $R$ is an isomorphism?

Question

Let $V$ be a finite-dimensional vector space, and let $R: V \rightarrow V$ be an invertible linear transformation.

Is there a linear transformation $T: V \rightarrow V$ such that $ \operatorname{Ker}TR = \{0\}$ but $ \operatorname{Ker}T \not= \{0\}$?

What I'd like you to help me with:

1. As I understand it (and my understanding is probably wrong), $R$ defines an isomorphism between $V$ to $V$ itself; But any vector space is isomorphic with itself, so it seems like a trivial thing...
2. I need some clues on how to approach the proof. I guess it should involve using dimensions, but I don't know how to start applying the relevant dimensions theorems

Answer 1

It is impossible to find such $T$, suppose that $T(x)=0$, there exists $y$ such that $R(y)=x$, $TR(y)=T(x)=0$