Is it consistent with ZFC that no nontrivial forcing notion has automatic mutual genericity? A nontrivial forcing notion $\newcommand\Q{\mathbb{Q}}\Q$ exhibits
automatic mutual genericity, if whenever $G,H\subseteq\Q$ are
distinct $V$-generic filters (existing, say, in some forcing
extension of $V$), then they are mutually generic.
In any forcing extension $V[G]$ by such forcing, $G$ must be the
only $V$-generic filter; so this kind of forcing leads to unique
generics. Thus, it is a rigidity concept. In particular, $\Q$
cannot be forcing equivalent below incompatible conditions, even
in a forcing extension by a filter containing one of them, since
then the isomorphic filter would be distinct but not mutually
generic with it.
Meanwhile, it is consistent with ZFC that there are forcing
notions $\Q$ with automatic mutual genericity.
For example, consider a Suslin tree $T$ that is
Suslin-off-the-generic-branch, which means that after forcing to
add a $V$-generic branch $g$ through $T$, then $T\upharpoonright
p$ remains Suslin in $V[g]$ for any node $p$ not on $g$. If $g$
and $h$ are two $V$-generic branches through $T$, then consider a
node $p$ on $h$ that is not on $g$. Since $T\upharpoonright p$ is
Suslin in $V[g]$, every maximal antichain in $V[g]$ is refined by
a level of the tree, and since $h$ goes through every level of the
tree, it follows that $h$ is $V[g]$-generic, and so they are
mutually generic. Gunter Fuchs and I discuss this property of Suslin trees in our
paper Degrees of rigidity for Suslin trees,
 J. Symbolic Logic, vol. 74, iss. 2, pp. 423-454, 2009. We prove
there that a $V$-generic Suslin tree is
Suslin-off-the-generic-branch in $V[T]$, and also one can
construct such trees from the $\Diamond$ principle.
Because all the examples of automatic mutual genericity of which I
am aware have the character of these Suslin tree examples, which
do not exist in every model of ZFC, I was wondering whether it might be possible that there are no forcing notions with automatic mutual genericity.
Question. Is it relatively consistent with ZFC that no
nontrivial forcing notion exhibits automatic mutual genericity?
Alternatively, I would be delighted if someone could exhibit in ZFC that there is a forcing notion with automatic mutual genericity.
This is question 10 of my recent paper, Upward closure and
amalgamation in the generic multiverse of a countable
model of set theory.
That paper is focused on various amalgamation and upward closure
results for countable models of set theory. For example, if $W$ is
any countable transitive model of set theory, then there are
$W$-generic Cohen reals $c$ and $d$ for which $W[c]$ and $W[d]$
have no common forcing extension. One can generalize the argument
(see the paper) to many other notions of forcing, including any
forcing notion $\Q$ that is wide, in the sense that $\Q$ is not
$|\Q|$-c.c. below any condition. I had wondered which other
forcing notions exhibit this non-amalgamation property, and by
means of the rigid Suslin tree examples was led to the concept of
automatic mutual genericity, which have amalgamation rather than
non-amalgamation.
 A: Omer Ben-Neria and I found out that Joel's example, and analogues at larger cardinals, are the only ones. Therefore, Joel's hunch that MA proves there are no c.c.c. forcings with automatic mutual genericity is correct, although it is not entirely clear to me how to give a positive answer to the question with no extra restrictions.
I'll sketch the proof here. First we showed that a poset $\mathbb{Q}$ satisfies automatic mutual genericity if and only if for every $p\perp q$ in $\mathbb{Q}$, forcing below $p$ does not add a new maximal antichain below $q$. This is the main part of the proof: one direction is trivial and the other involves collapsing $2^\mathbb{Q}$ to be countable and using a name for a new antichain $\dot{A}$ to construct $V$-generics $G,H$ below $p$ and $q$ which do not meet the dense-below-$(p,q)$ subset 
$$\{(p',q'):\exists q^* \textrm{ such that }q'\le q^*\textrm{ and }p'\Vdash q^*\in\dot{A}  \}.$$
Now suppose $\mathbb{Q}$ has automatic mutual genericity, and let $\kappa$ be least so that $\mathbb{Q}$ adds a new subset of $\kappa$. For simplicity, assume that $\mathbb{Q}$ is a complete Boolean algebra. If $\mathbb{Q}$ had an antichain $A$ of size $\ge\kappa$, then in the generic extension by $\mathbb{Q}$ we could take $p\in A$ and use the new subset of $\kappa$ to define a new antichain below $-p$. So $\mathbb{Q}$ is $\kappa$-c.c., and therefore it is a $\kappa$-Suslin algebra.
Now the characterization of posets with automatic mutual genericity implies that $\mathbb{Q}$ must remain "Suslin off the generic branch," since otherwise this is witnessed by a new antichain.
