Can we control where the critical point of an elementary embedding is mapped? At the risk of this question having an easy answer, I am asking the following:
Suppose we have a measurable cardinal $\kappa$ and a (possibly regular) cardinal $\lambda>\kappa$. Is there any condition, like the existence of an appropriate ultrafilter, that implies the existence of an elementary embedding $j:V\to M$ with critical point $\kappa$ so that $j(\kappa)=\lambda$?
An example is when $\kappa$ is huge: if there is a normal $\kappa$-complete ultrafilter $U$ on $P(\lambda)$ with the property that $\{X\subseteq P(\lambda)\mid ot(X)=\kappa\}\in U$, then the ultrapower embedding $j:V\to M\simeq Ult(V,U)$ satisfies $j(\kappa)=\lambda$ (and $~{}^{j(\kappa)}M\subseteq M$)..
However, I am interested in the case where $\kappa$ is only measurable, maybe not even strong. 
I am assuming Choice and whenever I write $j:V\to M$ I mean that $M\subseteq V$ is a transitive model of ZFC.
 A: Here is a necessary and sufficient criterion: $\lambda>2^\kappa$. This is easily seen to be necessary, since if $j:V\to M$ has critical point $\kappa$, then the power set $P(\kappa)$ is contained in $M$, and from this it follows that $j(\kappa)>2^\kappa$ since $j(\kappa)$ is inaccessible in $M$. Conversely, we can hit every cardinal above $2^\kappa$, by the following.
Theorem. If $\kappa$ is a measurable cardinal, then every cardinal $\lambda>2^\kappa$ is the image $\lambda=j(\kappa)$ of some elementary embedding $j:V\to M$ with critical point $\kappa$. 
Proof. Let $\mu$ be a normal measure on $\kappa$ and consider the class of images $j_\alpha(\kappa)$, where $j_\alpha$ is the $\alpha$-iterated ultrapower by $\mu$. These are the ordinals that form the critical sequence. This class of ordinals is closed and unbounded. It is clearly unbounded, since by iterating further, we can push $j(\kappa)$ as high as desired. It is closed, since the iterations are defined to take the direct limit at limit stages, and this makes the critical sequence continuous. 
Finally, I claim that every cardinal above $2^\kappa$ is on the critical sequence. To see this, it suffices to argue that at each stage, we don't jump over the next cardinal. If $j_\alpha:V\to M_\alpha$ is the $\alpha^{th}$ iterate, then the extender representation shows that every element of $M$ has the form $j(f)(s)$, where $f:\kappa\to V$ and $s$ is a finite sequence from the critical sequence below $\kappa_\alpha$, which are ordinals below $j_\alpha(\kappa)$. If we go one more step, to $j_{\alpha+1}:V\to M_{\alpha+1}$, then we only need to add one more generator or seed, namely $\kappa_\alpha$ itself, and so $|j_{\alpha+1}(\kappa)|^V\leq |\kappa^\kappa|\cdot|(\alpha+1)^{<\omega}|$, which has the same size as $j_\alpha(\kappa)$. So at successor stages, we don't get to the next cardinal, and so we reach all the cardinals at limit stages. So we'll get every cardinal above $2^\kappa$.
QED 
Lastly, note that although you had asked about cardinals $\lambda$ that are the images of $\kappa$ under an embedding, nevertheless such images are not always cardinals. For example, if $\mu$ is a measure on $\kappa$ with ultrapower $j:V\to M$, then $j(\kappa)$ is never a cardinal, since it is strictly between $2^\kappa$ and $(2^\kappa)^+$. So one might want a criterion for recognizing when an ordinal $\lambda$ is the image of $\kappa$ under an embedding, and I think that is a much subtler question.
