It is well-known that it is consistent with $ZF$ that the only automorphisms of the complex field $\mathbb{C}$ are the identity map and complex conjugation. For example, we have that $\vert\operatorname{Aut}(\mathbb{C})| = 2$ in $L(\mathbb{R})$. But suppose that we are given a nonprincipal ultrafilter $\mathcal{U}$ over the natural numbers $\mathbb{N}$. Is there any way to use $\mathcal{U}$ to define a third automorphism of $\mathbb{C}$?


Some background ... the "obvious" approach would be to note that the ultraproduct $\prod\_{\mathcal{U}} \bar{\mathbb{F}}\_{p}$ of the algebraic closures of the fields of prime order $p$ has lots of automorphisms arising as ultraproducts of Frobenius automorphisms. Of course, working in $ZFC$, this ultraproduct is isomorphic to $\mathbb{C}$ and hence we obtain many "strange " automorphisms of $\mathbb{C}$. However, the isomorphism makes heavy use of the Axiom of Choice and these fields are not isomorphic in $L(\mathbb{R})[\mathcal{U}]$. So a different approach is necessary if we are to find a third automorphism of $\mathbb{C}$ just in terms of $\mathcal{U}$ ...


Edit: Joel Hamkins has reminded me that I should mention that I always assume the existence of suitable large cardinals when I discuss properties of $L(\mathbb{R})$ and $L(\mathbb{R})[\mathcal{U}]$. For example, if $V = L$, then $L(\mathbb{R}) = L= V$ and so $L(\mathbb{R})$ is a model of $ZFC$. Of course, nobody would dream of studying $L(\mathbb{R})$ under the assumption that $V = L$ ...



Ultrafilters and automorphisms of the complex field
It seems not. 


It was shown by Di Prisco and Todorcevic (and reproved later by at least three sets of authors) that if sufficiently large cardinals exist (e.g., a proper class of Woodin cardinals), then after forcing with $\mathcal{P}(\omega)/\mathrm{Fin}$ (the infinite subsets of $\omega$, ordered by mod-finite containment) to produce a selective ultrafilter $U$, there is no selector (i.e., set meeting each equivalence class in exactly one point) for the equivalence relation $E\_{0}$ (mod-finite equivalence on $\mathcal{P}(\omega)$) in the inner model $L(\mathbb{R})[U]$.


It also seems to follow from ZF + DC$\_{\mathbb{R}}$ (which holds in $L(\mathbb{R})[U]$) that the existence of a discontinuous homomorphism from either of $(\mathbb{R}, +)$ or $(\mathbb{C}, +)$ to itself implies the existence of an $E\_{0}$ selector, as we will show below. Since a discontinuous automorphism of $(\mathbb{C}, +, \times)$ restricts to one for $(\mathbb{C}, +)$, this answers the question. 
The proof is the same for each of $(\mathbb{R}, +)$ and $(\mathbb{C}, +)$; moreover, the existence of each type of homomorphism implies the existence of the other. I haven't tried writing it up this way, but it seems that the argument can be carried out over an arbitrary complete additive metric group satisfying the triangle inequality. The existence of a discontinuous homomorphism of $(\mathbb{R}, +)$ easily gives one for
$(\mathbb{C}, +)$; we give a proof of the reverse direction at the end of this answer. 


So, let $h$ be a discontinuous homomorphism from $(\mathbb{R}, +)$ (or $(\mathbb{C}, +)$) to itself. 
As shown in the proof of Theorem 1 of a 1947 paper by Kestelman, for each positive real number $\delta$, $h$ is unbounded on $\{ x : |x| < \delta \}$. The same proof shows that the same fact holds for $(\mathbb{C}, +)$ (moreover, the fact follows easily from the definition of "discontinuous homomorphism"). Applying DC$\_{\mathbb{R}}$, we may find $\{ x\_{i} : i < \omega \}$ such that (1) each $|x\_{i}|$ is more than $\sum \{ |x\_{j}| : j > i\}$
and such that (2) for each $i$, $|h(x\_{i})| - \sum \{ |h(x\_{j})| : j < i \} > i.$


Let $X = \{ x\_{i}: i < \omega \}$ and let $Y$ be the set of reals (or complex numbers) which are sums of (finite or infinite) subsets of $X$ (note that all the infinite sums converge).
By condition (1) on $X$, each $y \in Y$ is equal to $\sum \{ x\_{i} : i \in S\_{y}\}$ for a unique subset $S\_{y}$ of $\omega$. 
Let $F$ be the equivalence relation on $Y$ where $y\_{0} F y\_{1}$ if and only if $S\_{y\_{0}}$ and $S\_{y\_{1}}$ have finite symmetric difference. 
By condition (2) on $X$, the $h$-preimage of each bounded subset of $\mathbb{R}$ ($\mathbb{C}$) intersects each $F$-equivalence class in only finitely many points (since if the bounded set is contained in an interval of length $i$, then for every $y$ in the intersection $S\_{y} \setminus i$ is the same, which can be seen be consideration of the maximum point of disagreement between the sets $S\_{y}$). It follows then that there is an $F$-selector : for each equivalence class, let $n \in \mathbb{Z}^{+}$ be minimal such that the $h$-preimage of $[-n, n]$ intersects the class, and then pick the least element of this intersection. Since $Y/F$ is isomorphic to $\mathcal{P}(\omega)/E\_{0}$ via the map $y \mapsto S\_{y}$, there is then an $E\_{0}$-selector.


As for getting a discontinuous homomorphism of $(\mathbb{R}, +)$ from one on $(\mathbb{C}, +)$ : Suppose that $h$ is a homomorphism of $(\mathbb{C}, +)$. Define $f\_{0},\ldots,f\_{3}$ on $\mathbb{R}$ as follows:
(1) If $h(x) = a + bi$, then $f\_{0}(x) = a$. 
(2) If $h(x) = a + bi$, then $f\_{1}(x) = b$. 
(3) If $h(iy) = a + bi$, then $f\_{2}(y) = a$.
(4) If $h(iy) = a + bi$, then $f\_{3}(y) = b$.
Then each of $f\_{0},\ldots,f\_{3}$ is a homomorphism of $(\mathbb{R}, +)$. 
Since $h(x + iy) = h(x) + h(iy) = f\_{0}(x) + if\_{1}(x) + f\_{2}(y) + if\_{3}(y),$ if all of $f\_{0},\ldots,f\_{3}$ are continuous 
then $h$ is. 


