$\DeclareMathOperator\hal{hal}$A field isomorphism $\phi:F\rightarrow G$ is a bijection such that (i) $\phi(x+y)=\phi(x)+\phi(y)$ and (ii) $\phi(xy)=\phi(x)\phi(y)$, where $F$ and $G$ are ordered fields. If $F=G$, then $\phi$ is called an automorphism. For $\mathbb{R}$, the only automorphism is the identity map. But for extensions of $\mathbb{R}$, like the hyperreals $^*\mathbb{R}$, there are many non-trivial automorphisms, if one assumes the Continuum Hypothesis.

(I understand that there are many hyperreal fields, as there are many ultrafilters to choose from. But for the purposes of this question, any arbitrary hyperreal field derived via the ultrapower construction with $\mathbb{R}$ embedded will do.)

Let $\hal(r)$ be the halo of $r$, where $r\in{^*\mathbb{R}}$ and is a real number. That is, $\hal(r)$ is the set of all $x\in{^*\mathbb{R}}$ such that $|x-r|<\frac{1}{n}$ for all $n\in\mathbb{N}$. Fix a $y\in \hal(r)$ such that $y>r$. For each $z$ such that $z\in \hal(r)$ and $z>y$, is there an automorphism $\phi$ such that the reals are fixed (i.e. for all real $q\in{^*\mathbb{R}}$, $\phi(q)=q$) and $\phi(y)=z$? Assume the Continuum Hypothesis.

Posted here but got no answer: https://math.stackexchange.com/questions/4937027/automorphism-on-the-hyperreals

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