Let $\mathfrak{g}$ be a finite dimensional Lie algebra over $\mathbb{R}$ and $\phi:\mathfrak{g}\to\mathfrak{g}$ be a Lie algebra automorphism. 

Viewing $\mathfrak{g}$ as a linear space and $\phi$ a linear automorphism, we can say $\phi$ is *hyperbolic* if the eigenvalues of $\phi$ are disjoint from $\lbrace z\in\mathbb{C}:|z|=1\rbrace$.

Then Proposition 3.6 in Smale's paper ([here][1]) says that:

 - Suppose that $\phi:\mathfrak{g}\to\mathfrak{g}$ is a Lie algebra automorphism which is hyperbolic as a linear map. Then $\mathfrak{g}$ must be nilpotent.

He also mentioned the following result in (Exercise in Bourbaki with hints: Algebras de Lie, Ex. 21b, p. 124.):

 - Let $\mathfrak{g}$ be a finite dimensional Lie algebra having an automorphism $\phi$, 
no eigenvalue of which is a root of unity, then $\mathfrak{g}$ is nilpotent.

Do you have ideas how to prove these results? 

Thanks!

  [1]: http://projecteuclid.org/euclid.bams/1183529092