Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers.

Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. (This uses the normality of $\mu$, and it is not necessarily true without that assumption; this was relevant for one of my dissertation results.)

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.

Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers.

Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. You can find this and related results in section 3 of my paper, Canonical seeds and Prikry trees, JSL 62, 1997 (adapted from a chapter of my dissertation). It uses the normality of $\mu$, and this particular fact is not necessarily true without that assumption, as shown in the paper.

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.

Yes, this situation can occur.

Suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull, the set $X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\}$, and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. (This uses the normality of $\mu$, and it is not necessarily true without that assumption; this was relevant for one of my dissertation results.)

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.

Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers.

Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. (This uses the normality of $\mu$, and it is not necessarily true without that assumption; this was relevant for one of my dissertation results.)

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.