**Notation and motivation.** Given an algebraic structure $\mathbb{M}$ of cardinality at most the continuum and with countably many operations, and a nonprincipal ultrafilter $\cal{U}$ on a countably infinite set, let $\mathbb{M}^{*}$ be the ultrapower of $\mathbb{M}$ modulo $\cal{U}$. It is well-known that if CH (continuum hypothesis) holds, then $\mathbb{M}^{*}$ is saturated (in the language of model theory) and therefore has many automorphisms.

Question.Let $\mathbb{Z}$ be the ring of integers, and $\mathbb{R}$ be the field of reals. Is it known if either $\mathbb{Z}^{*}$ or $\mathbb{R}^{*}$ has to have a nontrivial automorphism in the absence of CH?

**Remark.** The answer to the above question for the case of $\mathbb{Z}$ does not change if $\mathbb{Z}^{*}$ is replaced with $\mathbb{N}^{*}$ or with $\mathbb{Q}^{*}$, since automorphism groups are invariant under bi-interpretations, and:

(1) $\mathbb{Z}$ is bi-interpretable with $\mathbb{N}$ (thanks to Lagrange's four squares theorem), and

(2) $\mathbb{N}$ is bi-interpretable with $\mathbb{Q}$ (thanks to Julia Robinson's theorem).

In contrast, it is well-known that $\mathbb{C}^{*}$ has many automorphisms (independent of whether CH holds not) since $\mathbb{C}^{*}$ is isomorphic to $\mathbb{C}$ (by the Steinitz theorem that asserts that any two algebraically closed fields of the same uncountable cardinality are isomorphic), and the fact that the classical proof that $\mathbb{C}$ has $2^{2^{\aleph_0}}$ many automorphisms does not need CH in any of its steps.