$f$ is a normalized eigenform $\Rightarrow$ so is $f^\sigma$, for $\sigma \in \operatorname{Aut}\mathbb{C}$

Question

Let $f \in S_2(\Gamma_1(N))$ be a normalized eigenform, and $\sigma \in \operatorname{Aut}\mathbb{C}$. Then is $f^\sigma$ a normalized eigenform?

This is stated in theorem6.5.4. of Diamond, Shurman's "A first course in modular forms", with long proof. But I think, this is trivial by proposition 5.8.5.:

Let $f \in M_k(N, \chi)$. Then $f$ is a normalized eigenform if and only if these conditions hold: (1) $a_1 = 1$, and (2) $a_{p^r} = a_pa_{p^{r-1}} - \chi(p) p^{k-1} a_{p^{r-2}}$, and (3) $a_na_m = a_{nm}$ for coprime $n,m$.

Is my opinion wrong?

Thank you very much!