Nondegenerate involutions on the complex numbers — always just conjugation?

Suppose you have a ring homomorphism $(-)':\mathbb{C} \to \mathbb{C}$, which is an involution such that $\sum_i a_i a_i' = 0 \Leftrightarrow \forall i \ a_i=0$, where this sum is finite.

Must this be conjugate to complex conjugation, in the sense that there exists some bijection $\phi: \mathbb{C} \to \mathbb{C}$ such that $a' =\phi ^{-1} (\phi(a)^{\star})$, where $(-)^\star$ denotes complex conjugation?

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The sum is finite, countable, or what? In any case, a bijection $\phi$ with these properties exists if and only if the set of fixpoints of the involution and its complement have both cardinality $2^{\aleph_0}$. – Emil Jeřábek Oct 23 '12 at 13:39
I have edited the question to give some additional information: $(-)'$ is a ring homomorphism, and the sum must be finite. – Jamie Vicary Oct 23 '12 at 13:54

Yes. Your involution is a field automorphism, and since it is of order $2$, its fixed field $F$ is a subfield of $\mathbb C$ such that $[\mathbb C:F]=2$. (This incidentally implies that $F$ is a real-closed field, hence the condition on sums is automatically satisfied, you don’t have to assume it.) In particular, both $F$ and $\mathbb C\setminus F$ have cardinality $2^{\aleph_0}$, hence you can define $\phi$ as follows. Let $X$ be a selector on the equivalence relation given by $a\sim a'$ on $\mathbb C\setminus F$. Take any bijection of $F$ and $\mathbb R$, any bijection of $X$ and $\{z\in\mathbb C:\operatorname{Im}(z)>0\}$, and put $\phi(a)=\phi(a')^*$ for $a\in\mathbb C\setminus(F\cup X)$. You can make it slightly more constructive by observing that $\mathbb C=F(i)$, and $(a+ib)'=a-ib$ for $a,b\in F$. Thus, if $\phi$ is a bijection of $F$ and $\mathbb R$, we can extend it to an involution-preserving bijection $\mathbb C\to\mathbb C$ by putting $\phi(a+ib)=\phi(a)+i\phi(b)$.
We can also make $\phi$ preserve addition ($F$ is a vector space over $\mathbb Q$ of dimension $2^{\aleph_0}$, hence it is isomorphic to $\mathbb R$ as a vector space). However, we cannot in general make it preserve multiplication as well: $F$ is not necessarily isomorphic to $\mathbb R$ as a field, for example it may be non-archimedean. In fact, if $F$ is any real-closed field of cardinality $2^{\aleph_0}$, then $F(\sqrt{-1})$ is an algebraically closed field of characteristic 0 and transcendence degree $2^{\aleph_0}$, and as such it is isomorphic to $\mathbb C$. We may thus obtain an involutive field automorphism of $\mathbb C$ whose fixed field is isomorphic to $F$.