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.) 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_1(f)(\alpha_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$.
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.) 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_1(f)(\alpha_1))$. So $h$ is the identity, and since $h\circ j_0 = j_1$, we finally conclude that $j_0 = j_1$.
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.) 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$.
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not justonly for normal$\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete ultrafilters.
I needI'll do this by answering Trevor's question from the following theoremcomments:
Fact (UA). Assume UA, and letSuppose $U_0$ and $U_1$ beare countably complete ultrafilters with ultrapowerson ordinals $j_0 :V\to M_0$$\delta_0$ and $j_1 :V\to M_1$$\delta_1$. Then $j_1(U_0)$ belongs toLet $M_0$ if$j_0 :V\to M_0$ and only if there is an internal$j_1:V\to M_1$ denote their ultrapower embeddingembeddings, and assume $k : M_0\to M_1$ such that$j_{0}(P(\delta_0)) = j_{1}(P(\delta_1))$. Then $k\circ j_0 = j_1$$j_0 = j_1$. (The reverse direction is obvious
The fact suffices, since if $j_1(U_0) = k(j_0(U_0)).$)$j_0(U_0) = j_1(U_1)$, 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.
Now assumeProof of fact. Apply UA. Suppose to obtain internal ultrapower embeddings $U_0$$i_0 : M_0\to N$ and $U_1$ are ultrafilters,$i_1:M_1\to N$ such that $j_0 :V\to M_0$ and$i_0\circ j_0 = i_1\circ j_1$. Let $j_1:V\to M_1$ are their ultrapowers$\alpha_0 = [\text{id}]_{U_0}$, $\alpha_1 = [\text{id}]_{U_1}$, and $j_0(U_0) = j_1(U_1)$$\delta_* = j_0(\delta_0) = j_1(\delta_1)$. I'll show 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. AssumeSo assume without loss of generality that $U_0$ and$i_0(\alpha_0) < i_1(\alpha_1)$. Let $U_1$ are uniform ultrafilters$D$ be the $M_1$-ultrafilter on cardinals. Under UA$\alpha_1$ 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 constructibility preorder on such ultrafiltersfactor 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 linearan internal ultrapower embedding, one can conclude that (in fact$k_0$ is too. Since $D$ is an ultrafilter on an ordinal less than $\delta_*$, $D$ is coded in $M_1$ by a prewellordersubset of $\delta_*$ (using that $2^{<\delta_*} = \delta^*$ in $M_1$), so breaking symmetry, assume that $U_0\in L(U_1)$$D\in M_0$. ThenOne can show $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$$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 $A = (k_1\restriction \text{Ord})^{-1}[k_1(A)]\in M_0$. Therefore by the theoremUnder UA, $M_1\subseteq M_0$ implies that there is an internal ultrapower embedding $k : M_0\to M_1$$h : M_0\to M_1$ such that $k\circ j_0 = j_1$$h\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$(See Corollary 5. Recall that the Rudin-Keisler order is wellfounded and note:4.21 $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ Byhere.) From the elementarityperspective of $j_0$$M_0$, the Rudin-Keisler rank ofembedding $U_1$ is less than or equal to that$h$ preserves the powerset of $U_0$$\delta_*$. SinceBut $U_0\leq_\text{RK} U_1$$h(\delta_*) = h(j_0(\delta_0)) = j_1(\delta_0)\leq j_1(\delta_1) = \delta_*$, we must haveso by the Kunen inconsistency theorem, $U_0\equiv_\text{RK} U_1$$j\restriction\delta_*$ is the identity. In other wordsTherefore $h$ is surjective: if $a\in M_1$, $j_0 = j_1$$a = j_1(f)(\alpha_0) = h(j_1(f)(\alpha_1))$. So $h$ is the identity, as desiredand since $h\circ j_0 = j_1$, we finally conclude that $j_0 = j_1$.
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not just for normal ultrafilters, but for arbitrary countably complete ultrafilters.
I need the following theorem: Assume UA, and let $U_0$ and $U_1$ be countably complete ultrafilters with ultrapowers $j_0 :V\to M_0$ and $j_1 :V\to M_1$. Then $j_1(U_0)$ belongs to $M_0$ if and only if there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. (The reverse direction is obvious since $j_1(U_0) = k(j_0(U_0)).$)
Now assume UA. Suppose $U_0$ and $U_1$ are ultrafilters, $j_0 :V\to M_0$ and $j_1:V\to M_1$ are their ultrapowers, and $j_0(U_0) = j_1(U_1)$. I'll show $U_0 = U_1$. Assume without loss of generality that $U_0$ and $U_1$ are uniform ultrafilters on cardinals. Under UA, the constructibility preorder on such ultrafilters is linear (in fact, a prewellorder), so breaking symmetry, assume that $U_0\in L(U_1)$. Then $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$. Therefore by the theorem, there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$. Recall that the Rudin-Keisler order is wellfounded and note: $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ By the elementarity of $j_0$, the Rudin-Keisler rank of $U_1$ is less than or equal to that of $U_0$. Since $U_0\leq_\text{RK} U_1$, we must have $U_0\equiv_\text{RK} U_1$. In other words, $j_0 = j_1$, as desired.
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.
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.) 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_1(f)(\alpha_1))$. So $h$ is the identity, and since $h\circ j_0 = j_1$, we finally conclude that $j_0 = j_1$.
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not just for normal ultrafilters, but 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((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$$$\sigma_{\alpha,\kappa}((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$ 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 need the following theorem: Assume UA, and let $U_0$ and $U_1$ be countably complete ultrafilters with ultrapowers $j_0 :V\to M_0$ and $j_1 :V\to M_1$. Then $j_1(U_0)$ belongs to $M_0$ if and only if there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. (The reverse direction is obvious since $j_1(U_0) = k(j_0(U_0)).$)
Now assume UA. Suppose $U_0$ and $U_1$ are ultrafilters, $j_0 :V\to M_0$ and $j_1:V\to M_1$ are their ultrapowers, and $j_0(U_0) = j_1(U_1)$. I'll show $U_0 = U_1$. Assume without loss of generality that $U_0$ and $U_1$ are uniform ultrafilters on cardinals. Under UA, the constructibility preorder on such ultrafilters is linear (in fact, a prewellorder), so breaking symmetry, assume that $U_0\in L(U_1)$. Then $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$. Therefore by the theorem, there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$. Recall that the Rudin-Keisler order is wellfounded and note: $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ By the elementarity of $j_0$, the Rudin-Keisler rank of $U_1$ is less than or equal to that of $U_0$. Since $U_0\leq_\text{RK} U_1$, we must have $U_0\equiv_\text{RK} U_1$. In other words, $j_0 = j_1$, as desired.
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not just for normal ultrafilters, but 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 $$\sigma_\alpha((p_\delta)_{\delta\in I}) = (p_\delta)_{\delta\in I\cap \alpha}{}^\frown(p^*_\delta)_{\delta\in I\setminus \alpha}$$ 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 need the following theorem: Assume UA, and let $U_0$ and $U_1$ be countably complete ultrafilters with ultrapowers $j_0 :V\to M_0$ and $j_1 :V\to M_1$. Then $j_1(U_0)$ belongs to $M_0$ if and only if there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. (The reverse direction is obvious since $j_1(U_0) = k(j_0(U_0)).$)
Now assume UA. Suppose $U_0$ and $U_1$ are ultrafilters, $j_0 :V\to M_0$ and $j_1:V\to M_1$ are their ultrapowers, and $j_0(U_0) = j_1(U_1)$. I'll show $U_0 = U_1$. Assume without loss of generality that $U_0$ and $U_1$ are uniform ultrafilters on cardinals. Under UA, the constructibility preorder on such ultrafilters is linear (in fact, a prewellorder), so breaking symmetry, assume that $U_0\in L(U_1)$. Then $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$. Therefore by the theorem, there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$. Recall that the Rudin-Keisler order is wellfounded and note: $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ By the elementarity of $j_0$, the Rudin-Keisler rank of $U_1$ is less than or equal to that of $U_0$. Since $U_0\leq_\text{RK} U_1$, we must have $U_0\equiv_\text{RK} U_1$. In other words, $j_0 = j_1$, as desired.
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, and not just for normal ultrafilters, but 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 need the following theorem: Assume UA, and let $U_0$ and $U_1$ be countably complete ultrafilters with ultrapowers $j_0 :V\to M_0$ and $j_1 :V\to M_1$. Then $j_1(U_0)$ belongs to $M_0$ if and only if there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. (The reverse direction is obvious since $j_1(U_0) = k(j_0(U_0)).$)
Now assume UA. Suppose $U_0$ and $U_1$ are ultrafilters, $j_0 :V\to M_0$ and $j_1:V\to M_1$ are their ultrapowers, and $j_0(U_0) = j_1(U_1)$. I'll show $U_0 = U_1$. Assume without loss of generality that $U_0$ and $U_1$ are uniform ultrafilters on cardinals. Under UA, the constructibility preorder on such ultrafilters is linear (in fact, a prewellorder), so breaking symmetry, assume that $U_0\in L(U_1)$. Then $j_1(U_0)\in L(j_1(U_1)) = L(j_0(U_0)) \subseteq M_0$. Therefore by the theorem, there is an internal ultrapower embedding $k : M_0\to M_1$ such that $k\circ j_0 = j_1$. In particular, this means $U_0\leq_\text{RK} U_1$. Recall that the Rudin-Keisler order is wellfounded and note: $$|j_0(U_1)|_\text{RK}^{M_0}\leq k(|j_0(U_1)|_\text{RK}^{M_0}) = |j_1(U_1)|_\text{RK}^{M_1} = |j_0(U_0)|_\text{RK}^{M_0}$$ By the elementarity of $j_0$, the Rudin-Keisler rank of $U_1$ is less than or equal to that of $U_0$. Since $U_0\leq_\text{RK} U_1$, we must have $U_0\equiv_\text{RK} U_1$. In other words, $j_0 = j_1$, as desired.