Assume GCH. Suppose $j: V\to M$ is an elementary embedding such that $\mathrm{crit}(j)=\kappa$, ${}^\kappa M \subset M$, $M\supset V\_{\kappa+2}$. We can assume $j$ is defined from some extender of length $\kappa^{++}$. Hence, in particular, we know $\kappa^{++}<j(\kappa) < \kappa^{+++}$.


For each $\delta<j(\kappa)$, let $U\_\delta$ be an ultrafilter on $\kappa$ be defined such that $A\in U\_\delta$ iff $\delta\in j(A)$. Is it true that for any $\delta <\delta' < j(\kappa)$, $U\_{\delta}\neq U\_{\delta'}$? It is true if $\delta<\kappa^{++}$ since $\mathrm{Ult}(V, U\_\delta)$, $[id]\_{U\_\delta}=\delta$. Of course if the length of the extender is too long then this is not true since the number of ultrafilters on $\kappa$ is $\kappa^{++}$. In particular with appropriate hypothesis, it is possible that $j=j\_E$ where $E$ is some extender on $[\delta]^{<\omega}$ where $\delta\in \kappa^{+++}\cap \mathrm{cf}(\kappa^{++})$ and some $\alpha\neq\alpha'<j(\kappa)$ satisfy that $U\_\alpha=U\_{\alpha'}$. But $\delta$ may not be $\kappa^{++}$.



Number of ultrafilters in an extender
TLDR: Yes, all the ultrafilters are different.


Suppose $M$ is a transitive class and $j : V\to M$ is an elementary embedding. 


*Some notation.* 


* For any $x\in M$, let $H\_x = \{j(f)(x) : f\text{ is a function}\}\prec M.$
* Let $j\_x : V\to M\_x$ be the ultrapower by the derived ultrafilter $U\_x = \{A\subseteq X : x\in j(A)\}$ where $X$ is any set such that $x\in j(X)$.
* Let $k\_x : M\_x\to M$ be the unique elementary embedding with $k\_x\circ j\_x = j$ and $k\_x([\text{id}]\_{U\_x}) = x$.


Equivalently, $M\_x$ is the transitive collapse of $H\_x$, $k\_x : M\_x\to M$ is the inverse of the transitive collapse, and $j\_x = k\_x^{-1}\circ j$.


*A nonstandard definition.* An ordinal $\xi$ is a *weak generator* of $j$ if $\xi\neq j(f)(\alpha)$ for any function $f$ and any $\alpha < \xi$. (The difference between a generator and a weak generator is that in the definition of a generator one requires that $\xi\neq j(f)(\vec \alpha)$ for any $\vec \alpha \in [\xi]^{<\omega}$. Thus a generator is the same thing as an additively indecomposable weak generator.)


*A useful fact about elementary embeddings.* Suppose $x\in M$ and $\xi$ is the least ordinal such that $x \in H\_\xi$. Then $$H\_x = H\_\xi$$ Thus $j\_x = j\_\xi$ and $k\_x = k\_\xi$. Moreover $\xi$ is a weak generator of $j$. We call $\xi$ the *weak generator associated to $x$.*


*Key point:* In your situation, for any distinct weak generators $\xi,\xi'$ of $j$, $j\_\xi\neq j\_{\xi'}$. (Also weak generators are the same thing as generators in your situation, but this is not really relevant.) Since $V\_{\kappa+2}\subseteq M$, if $\xi < (2^\kappa)^+$ is a weak generator of $j$, then $$\xi < ((2^{\kappa})^+)^{M\_\xi} = \min (\text{Ord}\setminus H\_\xi)$$ Hence $((2^{\kappa})^+)^{M\_\xi}$ is the least weak generator of $j$ above $\xi$. Therefore for distinct weak generators $\xi,\xi'$, $((2^{\kappa})^+)^{M\_\xi}\neq ((2^{\kappa})^+)^{M\_{\xi'}}$, so $M\_\xi \neq M\_{\xi'}$, and certainly $j\_\xi \neq j\_{\xi'}$.


*The answer.* Assume that for any distinct weak generators $\xi,\xi'$ of $j$, $j\_\xi\neq j\_{\xi'}$. Suppose $X$ is a set, suppose $x,x'\in j(X)$, and assume the derived ultrafilter $U\_x = \{A\subseteq X : x\in j(A)\}$ is equal to $U\_{x'} = \{A\subseteq X : x'\in j(A)\}$. We will show $x = x'$. Let $\xi$ and $\xi'$ be the weak generators associated to $x$ and $x'$ respectively. Since $U\_x = U\_{x'}$, the associated ultrapower embeddings $j\_x$ and $j\_{x'}$ are also equal. It follows that $j\_\xi = j\_x = j\_{x'} = j\_{\xi'}$. Thus by our assumption about weak generators, $\xi = \xi'$. In particular, $k\_x = k\_{\xi} = k\_{\xi'} = k\_{x'}$. Therefore $$x = k\_x([\text{id}]\_{U\_x}) = k\_{x'}([\text{id}]\_{U\_{x'}}) = x'$$ Hence $x = x'$, as desired.


