Atomless measure space without measure preserving isomorphisms Question: Could somebody give an example of a nontrivial atomless measure space without measure preserving isomorphisms (except for the identity)? 
Background: A measure preserving isomorphism on a measure space $(X,\Sigma,\mu)$ is a bijection $\phi$ such that
$$\forall A\in\Sigma:\mu(\phi^{-1}(A))=\mu(\phi(A))=\mu(A)$$
Edit: By 'except for the identity' I obviously meant to exclude all $\phi$ such that $\mu(A\triangle\phi(A))=0$ for all $A\in\Sigma$. 
 A: Didn't know the answer to the original version, but only because of stupid/boring examples. Here's an example for the modified version: An atomless measure space such that if $\phi$ is any measurable bijection with a measurable inverse, measure-preserving or not, then $\phi(A)=A$ for every measurable set $A$.
Let $<$ be a well-ordering of $[0,1]$ (so in particular $<$ has nothing to do with the standard order). By transfinite recursion we can "construct" a strictly increasing map $\kappa$ from $[0,1]$ to the class of infinite cardinals, for example by taking $\kappa(x_0)=\aleph_0$ if $x_0$ is the smallest element of $[0,1]$ and $$\kappa(x)=2^{\bigcup_{x'<x}\kappa(x')}\quad(x>x_0).$$Let $(S_x)_{x\in[0,1]}$ be a pairwise disjoint collection of sets with $$|S_x|=\kappa(x).$$Set $$X=\bigcup_{x\in[0,1]}S_x.$$For $E\subset[0,1]$ let $$A_E=\bigcup_{x\in E}S_x.$$ Say $A\subset X$ is measurable if and only if $A=A_E$ for some Lebesgue-measurable $E\subset[0,1]$, and define $$\mu(A_E)=m(E),$$where $m$ is Lebesgue measure. No atoms ($S_x$ has no non-trivial measurable subsets but it's not an atom because it has measure zero).
Say $\phi:X\to X$ is a measurable bijection with a measurable inverse. Then induction on $x$ shows that $\phi(S_x)=S_x$. First, since $\phi(S_x)$ is measurable it must be a union of $S_y$ for some set of $y$'s. But $$|S_y|>|S_x|\quad(y>x),$$hence $$\phi(S_x)\subset\bigcup_{y\le x}S_y.$$By induction we have $\phi(S_y)=S_y$ for all $y<x$. Hence $\phi(S_x)\subset S_x$. Similarly $\phi^{-1}(S_x)\subset S_x$, so $\phi(S_x)=S_x$.
And hence $\phi(A)=A$ for every measurable set $A$.

Not an answer to the original version because there do exist plenty of measure-preserving isomorphisms other than the identity, in fact such that $\phi(x)\ne x$ for every $x$.
