Relating ${\rm ad}(\phi(X))$ with $\phi \circ\,{\rm ad}(X) \circ \phi^{-1}$.

Question

The exercise is to prove that $\phi: {\frak g}\to {\frak g}$ is an automorphism if and only if ${\rm ad}(\phi(X)) = \phi\circ {\rm ad}(X)\circ \phi^{-1}$. Here $\frak g$ is a Lie algebra, of course.

Assuming that $\phi$ is an automorphism, I managed to check the equality, but I'm a bit confused about the other direction. If I assume that equality, then $\phi^{-1}$ already exists, great. I don't know how to check that $\phi$ is a morphism only with that information, though. Perhaps the exercise is poorly written (I didn't omit anything). Can someone help me? Hints are ok, too. Thanks.

Answer 1

If $\phi$ is an automorphism, the equality means $ad(\phi(X))\circ\phi=\phi\circ ad(X)$ take that at $Y$, $ad(\phi(X))(\phi(Y))=\phi(ad(X)(Y))$ which is equivalent to $[\phi(X),\phi(Y)]=\phi([X,Y])$.

Answer 2

For all $x,y \in \mathfrak{g}$ you have $$ \phi([x,y]) = (\phi \circ \mathrm{ad}(X))(y) = (\mathrm{ad}(\phi(x)) \circ \phi)(y) = [\phi(x),\phi(y)]. $$