Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection between A and B? In particular, is this true in the model obtained by adding $\omega_2$ Cohen (resp. random reals) over a model of CH?

$\begingroup$ In your last sentence, did you perhaps mean to consider the null case in the random extension and the meager case in the Cohen extension? $\endgroup$ – Miha Habič Oct 11 '13 at 0:48

$\begingroup$ No. If you add $\omega_2$ randoms to a model of CH, the randoms added constitute a Sierpinski set of size continuum. $\endgroup$ – Ashutosh Oct 12 '13 at 18:50
$\newcommand\continuum{\mathfrak{c}}$
Update. This proof strategy is hopeless, for the reasons explained in Miha's answer at mathoverflow.net/a/144538/1946, and the comments on it. But I'll leave it here for the record of this false attempt.
Original answer:
Your hypothesis follows from the continuum hypothesis, and more generally, from the assertion that the additivity of the null ideal is the continuum, which is consistent with the failure of CH.
To see this, suppose that the additivity of the null ideal is the continuum, which means that the union of fewer than continuum many sets of measure zero still always has measure zero, and in particular, every set of size less than the continuum has measure zero. Fix any two sets $A$ and $B$ that are not of measure zero. So they have size continuum. Let us now build a bijection between them $\pi:A\to B$, in such a way that $\pi$ takes every measure zero subset of $A$ to a measurezero subset of $B$ and conversely. We will construct $\pi$ as the union of a increasing chain of partial functions $\pi=\bigcup_{\alpha\lt\continuum}\pi_\alpha$, in a construction of length continuum, where $\pi_\alpha:A_\alpha\to B_\alpha$ is a bijection of measurezero subsets of $A$ and $B$, respectively. Enumerate the Borel measurezero sets as $D_\alpha$ for $\alpha\lt\continuum$. Suppose that $\pi_\alpha:A_\alpha\to B_\alpha$ is defined, and consider the set $D_\alpha$. First, we may extend $A_\alpha$ to $A_{\alpha+1}$ in such a way that $D_\alpha\cap A\subset A_{\alpha+1}$, by also adding a measure zero part of $B$ to $B_\alpha$, forming $B_{\alpha+1}$ and extending the bijection to $\pi_{\alpha+1}:A_{\alpha+1}\to B_{\alpha+1}$. (I am using that every nonmeasurezero set contains a measure zero set of size continuum, which I believe follows from our assumptions; please correct me if this is wrong. Edit: this is not correct, because of the possibility of Sierpinski and Luzin sets, which exist, as Miha pointed out, when the additivity numbers are large.) Similarly, for the target side, we may extend in such a way also that $D_\alpha\cap B\subset B_{\alpha+1}$. At limit stages of our construction, we take the union of the bijections constructed so far, and until the end of the constructin, this union domain and target will still have measure zero by the assumption that the additivity of the null ideal is the continuum.
The resulting map $\pi:A\to B$ is a bijection, since every point will arise as a singleton in some $D_\alpha$, thereby getting added to the domain and range at that stage. And every measure zero subset of $A$ will be covered by some $A\cap D_\alpha$ for some $\alpha$, which gets mapped to $B_{\alpha+1}$, which has measure zero on the $B$ side. And similarly for the measure zero subsets of $B$.
Thus, the construction resembles the familiar backandforth constructions, but at each stage, we have defined the bijection on only measure zero sets, which exhaust neither $A$ nor $B$. This gives time to diagonalize against the possible measure zero sets, since there are only continuum many Borel measure zero sets to consider.
So the answer is yes, this situation is consistent.
Meanwhile, if the uniformity number (the smallest size of the nonmeasure zero set) is less than the continuum, then your hypothesis clearly fails, since there will be nonmeasurezero sets of different cardinality, which can therefore not be bijective. So we seem to have trapped your statement between the assertions that the additivity number is big and the uniformity number is not small.
A very similar argument seems to work also in the case of category rather than measure, under the assumption that the additivity of the meager ideal is the continuum. (And one similarly needs in this case that every nonmeager set contains a meager set of size continuum.)

$\begingroup$ Upon reflection, I'm less sure about the issue I mention in the second paragraph concerning measurezero sets of size continuum, and so we should regard this as a proof idea rather than a proof. Is it consistent with the additivity number being the continuum that every nonmeasurezero set must contain a measure zero set of size continuum? $\endgroup$ – Joel David Hamkins Oct 9 '13 at 11:34

$\begingroup$ For example, does every set of reals contain a measurezero set of the same cardinality? If the original set has the perfect set property, then the answer is yes, since we can find suitable Cantor sets (of measure zero) inside any perfect set. $\endgroup$ – Joel David Hamkins Oct 9 '13 at 16:18

$\begingroup$ I have now asked these questions at mathoverflow.net/questions/144529/… $\endgroup$ – Joel David Hamkins Oct 10 '13 at 23:09

$\begingroup$ This proof strategy is hopeless, for the reasons explained in Miha's answer at mathoverflow.net/a/144538/1946, and the comments on it. $\endgroup$ – Joel David Hamkins Oct 11 '13 at 0:41

$\begingroup$ Notice that in the Cohen model, every non null set has size continuum ($ = \omega_2$) and has a null subset of size continuum so it is at least not outright false that the restrictions of the null ideal to any two non null sets are isomorphic in this model. $\endgroup$ – Ashutosh Oct 12 '13 at 19:09