TITLE: Can a parent and child node have the same type in a well-founded digraph tree?
QUESTION [10 upvotes]: $\newcommand\toward{\rightharpoonup}$It would help me to
understand something in a current research project if someone
could provide an example of directed graph $\langle
G,\toward\rangle$ with the following properties

The graph is connected.
Every node $x$ points at exactly one parent node $y$, via $x\toward y$.
Thus, the underlying relation is a tree.
The relation $\toward$ is well-founded, so that every nonempty
subset $X\subset G$ has an $\toward$-minimal element $x\in X$, meaning that $y \not\!\toward x$ for all $y\in
 X$. In the presence of DC, this is equivalent to saying that there
is no infinitely receding sequence $\cdots\toward x_3\toward x_2\toward x_1\toward
 x_0$.
Finally, the key part, there are two elements $a$ and $b$ with
$a\toward b$, such that $a$ and $b$ realize exactly the same type in
$\langle G,\toward\rangle$. That is, $G\models\varphi(a)\leftrightarrow\varphi(b)$ for any assertion $\varphi$ in the digraph language.

Can this happen? 
Note that there can be no automorphism mapping $b$ to $a$, for in this case, we could iterate it to produce a receding chain, contrary to the well-founded hypothesis.
Also, since $b$ is not minimal, neither can $a$ be minimal, and if there is a path from a minimal element to $a$ of length $k$, then there is one to $b$ of length $k+1$, and so if the types are the same, there is a path to $a$ of length $k+1$, and so on. So both $a$ and $b$ will have infinite rank with respect to the well-founded relation. And I think one can carry that reasoning much further.
In my actual application, there is a lot more structure, and I consider $a$ and $b$ to realize the same assertions in a much stronger language, one that includes second-order logic and more. But I realized I don't have a complete grasp even on this "easy" case. What are types like in these well-founded digraph trees?

REPLY [6 votes]: What about this? Consider a graph $G$ of size $\omega_1$ such that 

$G$ is connected,
For every node $a$ of rank $\alpha$, and every $\beta<\alpha$, $a$ has $\omega_1$-many children of rank $\beta$.
Every node has a parent, and there are nodes of all ranks below $\omega_1$.

Now I think an Ehrenfeucht-Fraisse game should show that - if I pick nodes $a$ and $b$ in $G$ with $a$ the parent of $b$ and $rk(b)>\omega^\omega$ (I'm not sure that's optimal) - the structures $(G, a)$ and $(G, b)$ are elementarily equivalent, so $a$ and $b$ have the same type in $G$. I'm pressed for time, so I can't write out the details - of course, if there's a mistake, here it is! - but this seems to work.
We can then replace $G$ with a countable subgraph $G_0$, by taking a countable elementary substructure of $G$ containing $a$ and $b$.