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.

partof the definition of "tree", which itself is part of the definition of a Suslin tree. Of course, it's not a significant one, we can always add it "by force". Yes, it makes things a bit more awkward when you do these things, because then the root goes somewhere and it must be compatible withsomething. But that's why we forego the root in the requirement. Just like how defining a regressive function will generally require it to be regressive for non-zero ordinals (or: where possible). $\endgroup$ – Asaf Karagila Jul 12 '18 at 10:25