If $V$ is a vector space over $\mathbb Q$ with $\operatorname{dim}(V)=3$. How we can prove that there is NO automorphism $\phi: V \rightarrow V$ such that $\phi^{-1}=2\phi$.
I tried: Let $\phi: V \rightarrow V$ is an automorphism such that $\phi^{-1}=2\phi$, then
let $v,u\in V$, then since $\phi^{-1}$ is also a homomorphism we have:
$\phi^{-1}(uv)=\phi^{-1}(u)\phi^{-1}(v)=2\phi(u).2\phi(v)=4\phi(u)\phi(v)$ On the other hand, $\phi^{-1}(uv)=2\phi(uv)=2\phi(u)\phi(v)$, so
$4\phi(u)\phi(v)=2\phi(u)\phi(v)$
implies
$\phi(u)\phi(v)=0$, for all $u,v \in V$, which means $\phi=0$.
I think there is something missing in my argument! I didn't need the dimension of $V$ !!
Any suggestion!?