Fix a cardinal $\lambda$$\newcommand{\cU}{\mathcal U}\newcommand{\cV}{\mathcal V}$. Consider the equivalence relation on $\beta\lambda$ given by $\cU\sim \cV$ when for all first-order structures $M$ we have $M^{\cU}\cong M^{\cV}$.


By considering the structure $(\lambda,A)\_{A\subseteq \lambda}$, one can show that the $\sim$-classes have at most $2^\lambda$ elements (arguing as [here](https://math.stackexchange.com/a/3777179/30222)).


If we consider the action of the symmetric group $S(\lambda)$ on $\beta\lambda$, it is easy to see that for each $\cU\in \beta\lambda$ and $f\in S(\lambda)$, we have $\cU\sim f(\cU)$.


Is the converse also true, namely, if $\cU\sim \cV$, are they necessarily conjugate by a permutation of $\lambda$? If not, is it true for $\lambda=\aleph\_0$ (or any $\lambda$)?




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Edit: Given that the answer is positive (for signature of size $2^\lambda$), it would also be interesting to know whether the equivalence still holds if we define $\cU\sim \cV$ when $M^{\cU}\cong M^{\cV}$ for all first-order structures $M$ whose signature is either


1. strictly smaller than $\lambda$, or
2. of size at most $\lambda$, or
3. strictly smaller than $2^{\lambda}$.



When do two ultrafilters yield isomorphic ultrapowers?
If one allows an arbitrary signature, the answer to this question is fairly well-known. Consider the structure $(\lambda,A)\_{A\subseteq \lambda}$. Let $(M,R\_A)\_{A\subseteq \lambda}$ be the common ultrapower by $\mathcal U$ and $\mathcal V$. Note that $A\in \mathcal U$ if and only if $[\text{id}]\_U \in R\_A$ by the definition of an ultrapower. Fix functions $f,g:\lambda\to \lambda$ such that $[f]\_\mathcal V = [\text{id}]\_\mathcal U$ and $[g]\_\mathcal U = [\text{id}]\_\mathcal V$. We have $f\_\*(\mathcal V) = \mathcal U$: for any $A\subseteq \lambda$,
$A\in \mathcal U$ if and only if $[\text{id}]\_\mathcal U\in R\_A$ or equivalently
$[f]\_\mathcal V\in R\_A$, which by definition means $\{\alpha < \lambda : f(\alpha) \in A\}\in \mathcal V$ or $f^{-1}[A]\in \mathcal V$; that is, $A\in f\_\*(\mathcal V)$.
Similarly $g\_\*(\mathcal U) = \mathcal V$. It
follows that $(g\circ f)\_\*(\mathcal V) = \mathcal U$.
A fundamental theorem (discovered independently by Rudin, Keisler, Blass Katetov, Frolík, and maybe others, proof below) states that for
any ultrafilter $\mathcal W$ over $X$ and any $h :X\to X$, if $h\_\*(\mathcal W) = \mathcal W$
then $[h]\_\mathcal W =[\text{id}]\_\mathcal W$
Therefore $[g\circ f]\_\mathcal V = [\text{id}]\_\mathcal V$. Hence there is a set $A\in \mathcal V$ such that $g\circ f\restriction A$ is the identity. In other words, $f$ is one-to-one on a set in $\mathcal V$. It is then not hard to modify $f$ on a null set to make it a permutation.


*Proof that $h\_\*(\mathcal W) = \mathcal W$ implies $[h]\_\mathcal W= [\text{id}]\_\mathcal W$:* Assume not, and so without loss of generality $h(x) \neq x$ for all $x\in X$. Consider the graph $G$ with vertex set $X$ and edge set $E = \{\{x,y\}\in [X]^2 : h(x) = y\}$. We claim $G$ is $3$-colorable. Any finite connected induced subgraph $H$ of $G$ with $n$ vertices contains at most $n$ edges (as $x\mapsto \{x,f(x)\}$ is a partial surjection), and hence contains at most one cycle. Therefore removing at most one edge of $H$ yields an acyclic and hence 2-colorable graph, and this easily implies $H$ is $3$-colorable. By compactness, $G$ is $3$-colorable. Therefore there is a partition $\{A\_0, A\_1, A\_2\}$ of $X$ such that $G\restriction A\_n$ is discrete for $n =0,1,2$. This means that $h^{-1}[A\_n]\cap A\_n\neq \emptyset$ for $n =0,1,2$. Therefore if $A\_n\in \mathcal W$, then $A\_n\notin h\_\*(\mathcal W) =\mathcal W$, contradiction.


The question for countably incomplete ultrafilters and finite signatures is far more interesting ~~and seems to be sensitive to set theoretic hypotheses. The answer seems to be yes assuming Woodin's HOD Conjecture. Details on request.~~


**Details:** I can actually answer the finite signature question positively without the HOD Conjecture, although my proof looks like overkill. I need a lemma.


**Lemma.** *Suppose $i,j : V\_\alpha\to N$ are elementary embeddings that are continuous at regular cardinals $\delta\_0 < \delta\_1 < \alpha$, and suppose there is a partition $\vec S$ of $\{\alpha < \delta\_1 : \text{cf}(\alpha) = \delta\_0\}$ into $\delta\_1$ stationary sets such that $i(\vec S) = j(\vec S)$. Then $i\restriction \delta\_1 = j\restriction \delta\_1$.*


Given this, we proceed as follows. Consider the structure $M = (V\_\alpha,\in,\delta\_0,\delta\_1,\vec S,f)$ where $\delta\_0 > \lambda$ is regular, $\delta\_1 \geq 2^\lambda$ is regular, $\vec S$ is a stationary partition as in the lemma, and $f$ is a surjection from the cardinal $2^\lambda$ onto $P(\lambda)$. The ultrapowers of this structure by $\mathcal U$ and $\mathcal V$ coincide, and so we can identify them. Let's say the ultrapower is $(N,E,d\_0,d\_1,\vec T,g)$. (The signature has one relation symbol, interpreted as $E$, and four constant symbols.)


Let $i,j:V\_\alpha \to N$ be the ultrapower embeddings associated to $\mathcal U$ and $\mathcal V$. These embeddings are continuous at $\delta\_0$ and $\delta\_1$ since these cardinals are regular and above the underlying set $\lambda$ of $\mathcal U$ and $\mathcal V$. Since $i(\vec S) = \vec T = j(\vec S)$, the hypotheses of the lemma are true, so $i\restriction \delta\_1 = j\restriction \delta\_1$. In particular, these embeddings agree on the ordinal $2^\lambda$.


Now $i[P(\lambda)] = i(f)[i[2^\lambda]] = g[i[2^\lambda]] = g[j[2^\lambda]] = j[P(\lambda)]$. Inverting the transitive collapse shows $i\restriction P(\lambda) = j\restriction P(\lambda)$. This suffices to run the easy argument from the arbitrary signature case.


*Proof of Lemma.* Let $d\_0 = i(\delta\_0)$, $d\_1 = i(\delta\_1)$, $\vec T = i(\vec S) = \langle T\_a : a < d\_1\rangle$. We run an argument due to Solovay to show that $j[\delta\_1]$ is equal to the set $\{a < d\_1 : T\_a\text{ meets every $\delta\_0$-club in $d\_1$}\}$. By symmetry (and since $j(\vec S) = \vec T$), we have the same characterization of $j[\delta\_1]$, and this proves $i[\delta\_1] = j[\delta\_1]$, which easily implies the lemma.


(For the record, a set $C\subseteq d\_1$ is $\delta\_0$-club if it is cofinal in $d\_1$ and any increasing $\delta\_0$-sequence of elements of $C$ has a supremum in $N$ and this supremum belongs to $C$. Two(?) examples of $\delta\_0$-clubs : $i[\delta\_1]$ and $j[\delta\_1]$.)


First assume $T\_a$ meets every $\delta\_0$-club in $d\_1$, and we will show $a\in i[\delta\_1]$. The point is that $T\_a$ meets $i[\delta\_1]$. Take $\xi < \lambda$ with $i(\xi)\in T\_a$. Note that $\text{cf}(\xi) = \delta\_0$, so there is some $\alpha < \lambda$ with $\xi \in S\_\alpha$. Hence $\xi \in i(S\_\alpha) = T\_{i(\alpha)}$. Since $\xi \in T\_a\cap T\_{i(\alpha)}$ and the $T\_b$ are pairwise disjoint, $a = i(\alpha)$.


Conversely let us show that $T\_{i(\alpha)}$ (i.e., $i(S\_\alpha)$) meets every $\delta\_0$-club in $\delta\_1$. Towards this fix such a $\delta\_0$-club $C$. Then the usual argument shows that $C\cap i[\delta\_1]$ is a $\delta\_0$-club in $d\_1$. This means $i^{-1}[C]$ is a $\delta\_0$-club in $\delta\_1$ in the usual sense. Therefore there is some $\xi \in S\_\alpha\cap i^{-1}[C]$. Now $i(\xi)\in i(S\_\alpha)\cap C$, so $i(S\_\alpha)$ meets $C$, as desired. This proves the lemma.


