Measure preserving transformation that makes two partitions independent I am looking for a reference for the following result.  I think it is pretty well known but I haven't found it written down anywhere.
Let $(X, \mathcal{B}, \mu)$ be a standard nonatomic measure space and let $\mathcal{P}, \mathcal{P'}$ be two finite measurable partitions of $X$.  Then there is a map $\varphi: X \to X$ which preserves the measure $\mu$ and such that the partitions $\mathcal{P}$ and $\varphi^{-1}\mathcal{P'}$ are independent with respect to $\mu$.
(Two partitions $\mathcal{P}$ and $\mathcal{Q}$ are independent with respect to $\mu$ if for any two cells $A \in \mathcal{P}$, $B \in \mathcal{Q}$, $\mu(A \cap B) = \mu(A)\mu(B)$.)
 A: Atomic counterexample
I think it is false.
Take $X= \{a,b\}$ a space with just two point, $\mu(\{a\})=1/2$ and $\mathcal{P}=\mathcal{P'}=\{\{a\},\{b\} \}$ the full partition.
There are two maps preserving the measure: the identity $id$ and $\phi$ which permute the two elements.
For this two $id^{-1} \mathcal{P'}=\phi^{-1} \mathcal{P'}=\mathcal{P'}=\mathcal{P}$.
You can check that $\mathcal{P}$ is not self independent since $$
\mu(\{a\} \cap \{a\})=1/2 \ne \mu(\{a\})^2 =1/4.
$$
Non atomic counterexample
Let's recall that an atomic measure verify that for every $A$ measurable with $\mu(A)>0$ there is $B \subset A$ measurable with $0< \mu(B) < \mu(A)$. I think this property is still not strong enough for this result to be true. The following property, let's call it $\star$, is stronger.
For every $A$ measurable with $\mu(A)>0$, for every $x \in [0,\mu(A)]$, there is a $B \subset A$ measurable with $\mu(B)=x$.
Let's check that there exist non atomic measure which does not have $\star$ property.
Take $X=\{0,1\}^{\mathbb{N}}$ and take $x \in ]0,1[$ a transcendental number. Take the measure which, for every $I \in \mathbb{N}$, give mass $x$ to the set of sequence having $0$ at $i$ position. Then every measurable set being made with intersection and complement of such set, their measure are polynomial in $x$ and cannot be $1/2$.
Then we can create a counterexample. Take $Y=X_1 \cup X_2$ two copy of the previous space and $\mathcal{P}=\mathcal{P'}=\{ X_1 , X_2\}$.
There should be a measurable set of measure $1/4$ inside $X_1$ which is not the case.
I hope I didn't make a mistake here. I think $\star$ property should be strong enough for your property to be true but I haven't demonstrate it yet.
