We say $S$ is a Suslin forest if adding a minimum to $S$ we have a Suslin tree. So a Suslin Forest is essentially a Suslin tree $S$ in which we drop the requirement for $S$ to have a single root.
Notice that every uncountable subset of a Suslin tree is a Suslin forest.


Given a Suslin forest $S$, I would like to know wether it is possible or impossible to define a function $f:S\to S$ such that for every $x,y\in S$ with $x<y$ we have $f(x)\perp f(y)$ (i.e. $f(x)\nleq f(y) \land f(x)\ngeq f(y))$ and a proof or reference to a proof.


We say $S$ is a non-special Aronszajn forest if adding a minimum to $S$ we have a non-special Aronszajn tree.


Would the same result holds for a $S$ non-special Aronszajn forest which is not necessarily a Suslin forest?


**EDIT:**
For S Suslin forest, $f(S)$ is still a Suslin forest. For $S$ non-special Aronszajn forest which is not necessarily Suslin forest, we would like to require also that $f(S)$ is still a non-special Aronszajn.



Is there a function from a Suslin tree to itself which send compatible elements to incompatible elements?
Yes, this can happen, and indeed it happens in a subforest of any given Souslin tree.


Let's start with an illustrative case. Sometimes people consider Suslin trees that are not necessarily
normal, and where for example, a sequence converging to a limit level
can have more than one bounding node at that level.


Let us consider such a Suslin tree $T$, where every node is a
binary splitting node, and furthermore, every limit node has a
partner limit node with the same predecessors. Such a tree would
arise, for example, from a normal $2$-splitting Suslin tree, simply
by omitting the nodes at limit levels, in effect making the two
successors of that limit node the new successors of the branch
below.


Now let $S$ consist of two copies of $T$. This is a Suslin forest. Notice that every element of
$S$ can be determined by a binary sequence of some successor
ordinal length $\alpha+1$. Namely, the first bit tells you which copy you are in; the middle bits tell you about branching; and the limit bits tell you which of the two limit nodes you're at.


Let $f:S\to S$ be the function arising by simply swapping the last
bit of the binary representing sequences.


It follows that if $x<y$ in $S$, then $f(x)\perp f(y)$, since only
the last bit is changed. In other words, $f(y)$ extends $x$, which is incomparable with $f(x)$. 


The idea works generally, if one is willing to pass to a subforest. 


**Theorem.** Every Suslin tree $T$ has a Suslin subforest $S\subset T$, such that there is a function $f:S\to S$ with $x<y\to f(x)\perp f(y)$. 


**Proof.** I claim that there is a Suslin subforest $S\subset T$ such that every node of $S$ has another node on the same level, with the same predecessors. One can pick $S$ inductively: pick two incomparable nodes to be the roots; for every node you've picked, pick two incomparable successor nodes; and at limits, pick two incomparable nodes to serve as upper bounds. (One could also pick more than two, if desired.) Alternatively one can prune $T$ above the two roots by removing the limit level or non-branching nodes. This defines the forest $S$, which is Suslin because any chain or antichain in $S$ would give rise to a chain or antichain in $T$. 


The key property of $S$ is that every node $x$ in $S$ has another node $f(x)$ on the same level as $x$ and with the same predecessors as $x$. This property is sufficient to know that $x<y$ implies $f(x)\perp f(y)$, because $x$ will be below $f(y)$, but at the same level as $f(x)$ and incomparable with $f(x)$. 
$\Box$


One can consider the same operation on the complete binary tree
$2^{<\omega}-\{\langle\rangle\}$, without the empty
sequence. If you flip the last bit of every sequence to its
opposite, then you have $x<y\to f(x)\perp f(y)$. The function $f$
above is in effect simply doing this to each condition in the
finite part above the largest limit level preceeding a given node.


