Let $V$ be a vector space, $S,T\in\operatorname{End}(V),\phi\in\operatorname{Aut}(V)$ such that $T=\phi S\phi^{-1}$.
Prove that a subspace $W\subseteq V$ is $S$-invariant iff $\phi(W)$ is $T$-invariant.
My proof: $S(W)\subseteq W\stackrel?\Rightarrow\phi(S(W))\subseteq\phi(W)\Leftrightarrow\phi(S(\phi^{-1}(\phi(W))))\subseteq\phi(W)\Leftrightarrow T(\phi(W))\subseteq\phi(W)$ by using the relation between $S,T$ and $\phi$.
However, there is one step in the proof (the first one) which only seems to be true in one direction. As a counterexample for the return direction, take $f:\mathbb R\to\mathbb R, x\mapsto x(x-1)(x+1)$ which yields for example $f([-1,0[)\subseteq f([0,\infty[)$ but $[-1,0[\not\subseteq[0,\infty[$. Of course, in the original proof we are dealing with linear operators and not functions in general. How can I use the linearity to prove $\phi(S(W))\subseteq\phi(W)\Rightarrow S(W)\subseteq W$? Or do I have to prove the other direction completely differently?
The first step is true in both directions (remember that $\phi$ is an automorphism). You could think of it this way: $S(W) \subset W \implies \phi (S (W)) \subset \phi(W)$ and $\phi(S(W)) \subset \phi(W) \implies S(W) = \phi^{-1} ( \phi(S(W)) \subset \phi^{-1} (\phi(W)) = W$.
Elaboration upon request.