Let $\kappa$ be a measurable cardinal, we say that $\mathcal{D}$ is a normal ultrafilter iff whenever $g\in\kappa^\kappa$ such that $g<_\mathcal{D} Id$, we have some $\alpha<\kappa$ such that $\{\gamma<\kappa | g(\gamma) = \alpha\}\in\mathcal{D}$.
Given $\kappa$ a measurable cardinal and some $M$ which is a transitive collapse of an ultrapower by some $\kappa$-complete ultrafilter, $\mathcal{U}$, on $\kappa$, and $j_\mathcal{U}$ the canonical embedding of $V$ into $M$, we can reconstruct a normal ultrafilter by: $\mathcal{D} = \{X\subseteq\kappa | \kappa\in j_\mathcal{U}(X)\}$.
Now, given a set $A\in\mathcal{U}$ then $\kappa\in j_\mathcal{U}(A)$ iff $[Id]_\mathcal{U} E [C_A]_\mathcal{U}$ iff $\{\alpha<\kappa | \alpha\in A\}\in\mathcal{U}$. (where $E$ is the relation on the ultraproduct, and $C_x$ is the constant function giving $x$ for every $\alpha$)
Now, clearly if $A\in\mathcal{U}$ then $\{\alpha<\kappa | \alpha\in A\}\in\mathcal{U}$ and therefore $\kappa\in j_\mathcal{U}(A)$ and therefore $A\in\mathcal{D}$.
But that means the ultraproduct was given by a normal ultrafilter, even though we took some general ultrafilter, and I was told by my prof. that not every ultrafilter can be normalized through a permutation on $\kappa$.
Where and what am I doing wrong here?