When can we switch the order of forcing iteration I am interested in when two forcing iterations are isomorphic (or at least add the same reals) when the order of the forcings is switched. I know that each forcing does not properly exist in the ground model, so switching the order of forcing also amounts to changing names for forcings.
As a concrete example consider the proof of MA + not CH. As I understand, the idea is to iterate over every ccc poset.  If we switch the order, surely things might change.  But what if the posets are not chosen randomly.  Is there a normal way to show that two orderings of iteration are equivalent?  
I can imagine that there are two cases: the successor and limit cases. In case it matters, the forcings I am interested are all proper, the support is countable. 
Thanks for any advice, even to references where something like this is done.  
 A: Here's something you may or may not find useful. There is a construction, due to Laver, called termspace forcing, which tries to approximate a poset in a forcing extension with a ground model poset. Specifically, if $\mathbb{P}$ is a poset and $\dot{\mathbb{Q}}$ is a $\mathbb{P}$-name for a poset, we let 
$$A(\mathbb{P},\dot{\mathbb{Q}})=\{\tau;\tau\text{ is a $\mathbb{P}$-name and } 
\mathbb{P} \Vdash\tau\in\dot{\mathbb{Q}}\}$$
and order it by $\tau\leq\sigma$ iff $1\Vdash\tau\leq\sigma$. The intuition here is that $A(\mathbb{P},\dot{\mathbb{Q}})$ is what we would want the projection of $\mathbb{P}*\dot{\mathbb{Q}}$ onto the second coordinate to be.
The key result concerning termspace forcing is that $\mathbb{P}*\dot{\mathbb{Q}}$ embeds into $A(\mathbb{P},\dot{\mathbb{Q}})\times \mathbb{P}$ (the actual result is stronger, in that the filters for $\mathbb{P}$ and the termspace forcing needn't be mutually generic to give a generic for the iteration). This means that, modulo some quotient forcing, we have basically switched around the order of the iteration. The devil, as always, is in the details; the quotient forcing is in general very badly behaved (e.g. even for $\mathbb{C}=\text{Add}(\omega,1)$, the poset $A(\mathbb{C},\mathbb{C})$ has antichains of size continuum).
There is some more information on termspace forcing in Cummings' chapter of the Handbook, but otherwise I've found it quite difficult to find a good source on it.
A: It seems that the question can be quite non-trivial, even for two-step iterations. Shelah proved (see section 9, "Poor Cohen commutes only with himself" in the link below) that if $\mathbb{Q}$ is a Suslin ccc forcing adding a non-Cohen real, then in $V^{\mathbb{Q}}$ the old reals are meagre, which implies (by another result in that chapter) that $\mathbb{Q}$ does not commute with Cohen.
The paper: http://www.heldermann-verlag.de/jaa/jaa10/jaa10006.pdf
