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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 $\operatorname{Aut}(\mathbb{C})| = 2$ 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 $V = L$ ...

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 $V = L$ ...

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 $|Aut(\mathbb{C})| = 2$ 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 $V = L$ ...

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 $\operatorname{Aut}(\mathbb{C})| = 2$ 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 $V = L$ ...

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 $|Aut(\mathbb{C}| = 2$ 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 $V = L$ ...

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 $|Aut(\mathbb{C})| = 2$ 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 $V = L$ ...