Why does conjugation by a diagonalizable matrix induce a semisimple automorphism?

Question

This is a follow up to a question I asked earlier (Why is conjugation by a diagonal matrix a semisimple automorphism of $\textrm{GL}_n$?). Let $G = \textrm{GL}_n$, $s \in G$ a diagonalizable matrix, and let $\phi: G \rightarrow G$ be the automorphism $x \mapsto sxs^{-1}$. Is it obvious that the comultiplication map $\phi^{\ast}: k[G] \rightarrow k[G]$ is semisimple? I.e. $k[G]$ is the union of finite dimensional $\phi^{\ast}$-invariant subspaces on which $\phi^{\ast}$ acts as a diagonalizable linear transformation?

Answer 1

Yes, since left and right multiplication by a semisimple element are so, and they are commutative.