The point is that the ultrapower of any structure $\mathcal{M}$ by
a nonprincipal ultrafilter $\mu$ on $\mathbb{N}$ is countably
saturated, that is, it realizes any $n$-type with a countable
number of parameters, by simple argument: if $p(x)$ is a 1-type
consistent with the diagram of the ultrapower of $\mathcal{M}$, a
structure in a countable language, by the ultrafilter $\mu$, and
$p(x)$ uses only countably many parameters from $\mathcal{M}$,
then for each $i$ let $x_i\in M$ witness the first $i$ many
instances of the type, so that the function $i\mapsto x_i$
realizes all the assertions of the type for all sufficiently large
$i$, and thus the function $\langle x_i\mid i\in\mathbb{N}\rangle$
realizes the whole type. So now by induction we realize all
consistent $n$-types.

It follows now from CH that if any two ultrapowers of a countable
structure $\mathcal{M}$ by $\mu_0$ and $\mu_1$ are both saturated
models of size $\omega_1$, with the same theory, then being
saturated, they are isomorphic by a back-and-forth argument. We
enumerate both of them in type $\omega_1$, and build up an
isomorphism in stages by mapping the next element of one side to
any element of the other realizing the same type over the elements
that have been determined in the isomorphism so far.

So the general conclusion---the content of the "obvious" claim
that you mention---is that under CH, there is up to isomorphism at
most one ultrapower of any countable structure in a countable
language by a nonprincipal ultrafilter on $\mathbb{N}$, and
indeed, at most one ultrapower up to isomorphism of any structure
in a countable language of size at most $\omega_1$.