I am looking for proof verification:
For the group algebra $\mathbb{C}^{G}$, let $\sigma$ be an automorphism of the group $G$; then $e_g \mapsto e_{\sigma(g)}$ induces an automorphism on $\mathbb{C}^{G}$.
Group Algebra: If $G$ is a group of order $n$, and $\left\{ e_g : g \in G \right\}$ be an orthonormal basis for $\mathbb{C}^n$, define $e_g * e_h := e_{gh}$, and extend the product to all other vectors by distributivity. The result is a Banach algebra $\mathbb{C}^G$ with unity $e_1$ and the 1-norm.
Let $\sigma$ be an automorphism where $\sigma : G \to G$. We need to show that the map $\varphi : \mathbb{C}^G \to \mathbb{C}^G$ induces an automorphism where $e_g \mapsto e_{\sigma(g)}$. So we need to show that $\varphi$ is an isomorphism. So take $e_g, e_h \in \mathbb{C}^G$ such that $\varphi(e_g) = \varphi(e_h)$ and then $e_{\sigma(g)} = e_{\sigma(h)}$, and because $\sigma$ is an automorphism we have that $e_g = e_h$, showing injectivity. Then, for any $e_g$ in the range there is a corresponding $e_{\sigma(h)} = e_g$ as $\sigma$ is an automorphism, which shows that $\varphi$ is surjective and thus an isomorphism.