I asked this question on Mathematics Stackexchange but got no answer.
Question. Does $ZFC$ prove that there are non-principal ultrafilters $\mathcal U$ and $\mathcal V$ over $\mathbb N$ such that the respective ultrapowers $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are not isomorphic when regarded as extensions of the field $\mathbb R$?
$ZFC+\neg CH$ proves that such ultrafilters do exist. More precisely, $ZFC+\neg CH$ proves the stronger statement that there are $\mathcal U$ and $\mathcal V$ as above such that $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are not even isomorphic as fields. On the other hand, $ZFC+CH$ proves that $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ are always isomorphic as fields.
The question could be asked with $ZFC+CH$ instead of $ZFC$, but, seeing no reason to think that it cannot be settled in $ZFC$, the above formulation seemed more natural to me.
Since the fields $\mathbb R^{\mathcal U}$ and $\mathbb R^{\mathcal V}$ have $2^{2^{\aleph_0}}$ distinct structures of extensions of $\mathbb R$, as shown by Eric Wofsey, a field isomorphism between these two fields will not be $\mathbb R$-linear in general.
The transcendence degree of $\mathbb R^{\mathcal U}$ over $\mathbb R$ is $2^{\aleph_0}$. [Of course the same holds for $\mathbb R^{\mathcal V}$.] Indeed, for $x\in\mathbb R$ define $f_x\in\mathbb R^{\mathbb N}$ by $f_x(n)=e^{e^{nx}}$ [or $f_x(n)=\exp(\exp(nx))$ if you don't like small letters] and let $g_x$ be the image of $f_x$ in $\mathbb R^{\mathcal U}$. Then a straightforward argument shows that the subset $\{g_x\ |\ x>0\}\subset\mathbb R^{\mathcal U}$ is algebraically independent over $\mathbb R$.
As shown by tomasz, $ZFC$ proves that there are models $\mathcal A$ and ultrafilters $\mathcal U$ and $\mathcal V$ as above such that $\mathcal A^{\mathcal U}$ and $\mathcal A^{\mathcal V}$ are not isomorphic.
