Let $L$ be a finite dimensional semisimple Lie algebra over $\mathbb{C}$ and $H$ a maximal toral. Let $\Phi$ be the root system of $L$ relative to $H$. Then $$L=H\oplus (\oplus_{\alpha\in\Phi} \, L_{\alpha} ).$$ Let $\sigma$ be an automorphism of Lie algebra $L$ such that $\sigma(H)=H$. Since the components $L_{\alpha}$ in above decomposition are uniquely determined by $H$, it (intuitively) follows that $\sigma$ should permute the components $L_{\alpha}$'s, and consequently $\alpha$'s. My question is, in what precise way this permutation is? I confused with my calculations:
$$L_{\alpha}:=^{def} \{ x\in L \, | \, [h,x]=\alpha(h)x \,\, \forall h\in H\}.$$ So if $x\in L_{\alpha}$ then $[h,x]=\alpha(h)x$ for all $h\in H$. Applying $\sigma$, we get $$[\sigma(h),\sigma(x)]=\alpha(h).\sigma(x)\,\,\forall h\in H.$$ Replace $h$ by $\sigma^{-1}h$, and as $h$ varies over all $H$, $\sigma^{-1}(h)$ will vary over all $H$, so we get $$[h,\sigma(x)]=\alpha\sigma^{-1}(h)\sigma(x).$$ This implies $\sigma(x)$ is in $L_{\alpha\sigma^{-1}}$. So $\sigma(L_{\alpha})=L_{\alpha\sigma^{-1}}$; is this correct?
Yes, that is correct. Risking some abuse of notation, you can define that as an action of $\sigma$ on the roots, setting
$^\sigma \alpha := \alpha \circ \sigma^{-1}$
The reason why this is a bit more complicated than one might naively expect is that the roots are elements of the dual space $H^*$of $H$. In a way, what we're looking at here is the contragredient action to the original action of $\sigma$ on $H$.