TITLE: Do field automorphisms of a character imply outer automorphisms of the group?
QUESTION [7 upvotes]: Apologies for the imprecise wording of the title.
In studying the basic representation theory of finite groups, I've been struck by a pair of phenomena present in every example I've worked with but not mentioned in any texts I've studied. I am wondering if either or both of these phenomena is true in general, or perhaps true under some assumptions that are usually satisfied for small groups.
Phenomenon #1: We are given a finite group $G$ and a representation $\rho:G\rightarrow GL(V)$ on a complex vector space $V$, with character $\chi$.  Suppose $\chi(a)$ is an algebraic irrational, say $\alpha$, for some conjugacy class $a$ of $G$.  Then for any conjugate $\alpha'$ of $\alpha$ (i.e. another root of $\alpha$'s minimal polynomial over $\mathbb{Q}$), there exists a conjugacy class $a'$ of $G$, of the same size as $a$, such that $\chi(a')=\alpha'$.
Phenomenon #2: Furthermore, there is an outer automorphism of $G$ that carries $a$ to $a'$.

Are these patterns at all general? If so, what assumptions are needed for them to hold?  If not, what are some counterexamples?

Thanks so much!

REPLY [5 votes]: #1 is true.  If $a$ has order $n$, then by conjugation, we may assume that $\rho(a)$ is diagonal with $n$th roots of unity on the diagonal. Thus $\alpha = \chi(a)$ is a sum of $n$th roots of unity.  If $\alpha'$ is a conjugate of $\alpha$, then $\sigma(\alpha) = \alpha'$ for some $\sigma \in$ Gal($\mathbb{Q}(\mu)/\mathbb{Q}$) (where $\mu$ is a primitive $n$th root of unity).  We must have $\sigma(\mu) = \mu^k$ for some $k$ prime to $n$, in which case $\sigma$ sends all $n$th roots of unity to their $k$th powers.  Then $\rho(a^k)$ is also diagonal with all entries from $\rho(a)$ raised to the $k$th power. Hence $\chi(a^k) = \alpha'$.  
The conjugacy classes of $a$ and $a^k$ have the same size because the $k$th power map and the $k^{-1}$ (mod $n$) power map are inverse maps between the two conjugacy classes.
#2 seems too good to be true, although I don't have a counterexample at hand.  It would imply that any group having a non-rational character would have an outer automorphism.