Here, $\, j\_U, \, j\_D$ are the canonical elementary embeddings induced by $U,D$ respectively.


I note that it is consistent with the existence of a measurable that the answer be *yes*: it is true in the model $L[D]$ for $D$ a measure on $\kappa$.



If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U(U) = j_D(D)$, is $U=D$?
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete ultrafilters.


**First I'll show that in the Kunen-Paris model, there exist distinct normal ultrafilters $U\_0$ and $U\_1$ such that $j\_{U\_0}(U\_0) = j\_{U\_1}(U\_1)$. Moreover, $M\_{U\_0} = M\_{U\_1}$.**


Let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let $j : V\to M$ denote its ultrapower. Assume $2^\kappa = \kappa^+$. Let $\mathbb P$ be the Easton product $\prod\_{\delta\in I}\text{Add}(\delta,1)$ where $I\subseteq\kappa$ is a $U$-null set of cardinals.
Let $\mathbb Q = j(\mathbb P)$ and let $\mathbb Q/\mathbb P$ denote the product $\prod\_{\delta\in j(I)\setminus \kappa}\text{Add}(\delta,1)$ as computed in $M$. Thus $\mathbb Q \cong \mathbb P\times (\mathbb Q/\mathbb P)$. Since $\kappa\notin j(I)$, $\mathbb Q/\mathbb P$ is ${\leq}\kappa$-closed and $j(\kappa)$-cc in $M$, so by standard arguments, one can construct an $M$-generic filter $G\subseteq \mathbb Q/\mathbb P$ in $V$.


Let $H\subseteq \mathbb P$ be a $V$-generic filter. Let $j\_0:V[H]\to M[H\times G]$ be the unique lift of $j$ such that $j\_0(H) = H\times G$. Let $\sigma\_{\alpha,\kappa}$ denote the automorphism of $\mathbb P$ given by $$\sigma\_{\alpha,\kappa}((p\_\delta)\_{\delta\in I}) = (p\_\delta)\_{\delta\in I\cap \alpha}{}^\frown(p^\*\_\delta)\_{\delta\in I\setminus \alpha}$$
where for $q\in \text{Add}(\delta,1)$, $q^\*$ denotes the result of flipping the bits in $q$. Denote the similar automorphism of $\mathbb Q$ by $\sigma\_{\alpha,j(\kappa)}$. Let $j\_1 : V[H]\to M[H\times G]$ be the lift of $j$ such that $j\_1(H) = \sigma\_{\kappa,j(\kappa)}(H\times G)$.


Now it's time to show $j\_0(j\_0) = j\_1(j\_1)$. Since $j\_0(j\_0)\restriction M = j\_1(j\_1)\restriction M$, it suffices to show that $j\_0(j\_0)(H\times G) = j\_1(j\_1)(H\times G).$ This follows from a long fun computation:


\begin{align\*} 
 j\_0(j\_0)(H\times G) 
 &= j\_0(j\_0)(j\_0(H))\\
 &= j\_0(j\_0(H))\\
 &= j\_0(H\times G)\\
 &= j\_0(H)\times j(G)\\ 
 &= H\times G\times j(G)\\
 &= \sigma\_{\kappa,j(j(\kappa))} \circ \sigma\_{\kappa,j(j(\kappa))}(H\times G\times j(G))\\
 &= \sigma\_{\kappa,j(j(\kappa))}\circ \sigma\_{j(\kappa),j(j(\kappa))} (\sigma\_{\kappa,j(\kappa)}(H\times G)\times j(G))\\
 &= \sigma\_{\kappa,j(j(\kappa))}\circ \sigma\_{j(\kappa),j(j(\kappa))} (j\_1(H\times G))\\
 &= \sigma\_{\kappa,j(j(\kappa))} (j\_1(\sigma\_{\kappa,j(\kappa)}(H\times G)))\\
 &= \sigma\_{\kappa,j(j(\kappa))}(j\_1(j\_1(H)))\\ 
 &= \sigma\_{\kappa,j(j(\kappa))}(j\_1(j\_1)(j\_1(H)))\\
 &= j\_1(j\_1)(\sigma\_{\kappa,j(\kappa)}(j\_1(H)))\\
 &= j\_1(j\_1)(j\_0(H))\\
 &= j\_1(j\_1)(H\times G)
\end{align\*}


Finally, let $U\_0$ and $U\_1$ be the normal ultrafilters derived from $j\_0$ and $j\_1$. Since $j\_0(j\_0) = j\_1(j\_1)$, $j\_0(U\_0) = j\_1(U\_1)$, as desired.


**Second I'll sketch a proof that under the Ultrapower Axiom, the answer to your question is yes for arbitrary countably complete ultrafilters.**


I'll do this by answering Trevor's question from the comments:


**Fact** (UA). *Suppose $U\_0$ and $U\_1$ are countably complete ultrafilters on ordinals $\delta\_0$ and $\delta\_1$. Let $j\_0 :V\to M\_0$ and $j\_1:V\to M\_1$ denote their ultrapower embeddings, and assume $j\_{0}(P(\delta\_0)) = j\_{1}(P(\delta\_1))$. Then $j\_0 = j\_1$.*


The fact suffices, since if $j\_0(U\_0) = j\_1(U\_1)$, the hypotheses of the fact hold, and hence $j\_0 = j\_1$, which means $j\_0(U\_0) = j\_0(U\_1)$, so $U\_0 = U\_1$. I'll sketch a direct proof of the fact assuming $2^{{<}\delta\_0} = \delta\_0$, although with significantly more work, one can do without.


*Proof of fact.* Apply UA to obtain internal ultrapower embeddings $i\_0 : M\_0\to N$ and $i\_1:M\_1\to N$ such that $i\_0\circ j\_0 = i\_1\circ j\_1$. Let $\alpha\_0 = [\text{id}]\_{U\_0}$, $\alpha\_1 = [\text{id}]\_{U\_1}$, and $\delta\_\* = j\_0(\delta\_0) = j\_1(\delta\_1)$.
Note that $U\_0$ is the ultrafilter derived from $i\_0\circ j\_0$ using $i\_0(\alpha\_0)$ and likewise for $U\_1$, so if $i\_0(\alpha\_0) = i\_1(\alpha\_1)$, then $U\_0 = U\_1$, and we're done. So assume without loss of generality that $i\_0(\alpha\_0) < i\_1(\alpha\_1)$. Let $D$ be the $M\_1$-ultrafilter on $\alpha\_1$ derived from $i\_1$ using $i\_0(\alpha\_0)$, let $k\_1 : M\_1\to P$ be its ultrapower, and let $\ell:P \to N$ be the factor map. One can define $k\_0 : M\_0\to P$ by $k\_0([f]\_{U\_0}) = [j\_1(f)]\_D$. Since $\ell\circ k\_0 = i\_0$ is an internal ultrapower embedding, one can conclude that $k\_0$ is too. Since $D$ is an ultrafilter on an ordinal less than $\delta\_\*$, $D$ is coded in $M\_1$ by a subset of $\delta\_\*$ (using that $2^{<\delta\_\*} = \delta^\*$ in $M\_1$), so $D\in M\_0$. One can show $j\_D^{M\_0}\restriction \text{Ord} = k\_1\restriction \text{Ord}$, and as a consequence $M\_1\subseteq M\_0$: if $A$ is a set of ordinals in $M\_1$, then $k\_1(A)\in P\subseteq M\_0$, so $k\_1(A)\in M\_0$, so $A = (k\_1\restriction \text{Ord})^{-1}[k\_1(A)]\in M\_0$. Under UA, $M\_1\subseteq M\_0$ implies that there is an internal ultrapower embedding $h : M\_0\to M\_1$ such that $h\circ j\_0 = j\_1$. (See Corollary 5.4.21 [here](https://arxiv.org/pdf/2006.03293.pdf).) From the perspective of $M\_0$, the embedding $h$ preserves the powerset of $\delta\_\*$. But $h(\delta\_\*) = h(j\_0(\delta\_0)) = j\_1(\delta\_0)\leq j\_1(\delta\_1) = \delta\_\*$, so by the Kunen inconsistency theorem, $j\restriction\delta\_\*$ is the identity. Therefore $h$ is surjective: if $a\in M\_1$, $a = j\_1(f)(\alpha\_0) = h(j\_0(f)(\alpha\_1))$. So $h$ is the identity, and since $h\circ j\_0 = j\_1$, we finally conclude that $j\_0 = j\_1$.


