{\rm and} \ c'(C') = c(C).$$ We now compare the involutions $c\tau c^{-1}$ and $c'\sigma c'^{-1}$ of $G$: they both leave the torus $S$ invariant, and along with it, its Lie algebra and the chamber $c(C)$ inside it: both are realized in terms of permuting the simple roots that determine the chamber, the permutation being necessarily a Dynkin diagram automorphism.

Since $G$ is different from ${\rm Spin}(8)$, there is a unique such (involutive) permutation which is not the identity map. But none of the automorphisms $c\tau c^{-1}$ and $c'\sigma c'^{-1}$ is inner, thus by composing them and then restricting the result to the chamber $c(C)$, one gets the identity map; as an automorphism of $G$, this composition must consequently be an inner automorphism $c_g$ defined by some $g\in S$. That is, $$c\tau c^{-1}=c'\sigma c'^{-1} c_g.$$ Notice that the automorphism in the left hand side of the previous equation leaves $c(T_K)\subset S$ pointwise fixed. Consequently, $c'\sigma c'^{-1}$ does the same. In other words, $c(T_K)$ is contained in the fixed point set of $c'\sigma c'^{-1}$, which is just $c'(H)$. This finishes the proof.

[*Proof of Thm. \[main\] in the case $G$ simple.*]{} First assume that $G$ is different from ${\rm Spin}(8)$. By 