Automorphisms of locally finite countable posets-2 Given is a locally finite countable connected poset which satisfies further the following properties: 


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*Let $C$ be any maximal chain  ( i.e. inextendible chain) and $A$ be any antichain. Then $A$ is covered by both the sets ${\rm Past}(x)$ and ${\rm Future}(x)$  for $x$ running over $C$ i.e. $A \subset \bigcup_{x \in C} ({\rm Past}(x) \cup {\rm Future}(x))$ and where ${\rm Past}(x):=\lbrace y\mid y \leq x\rbrace$ and ${\rm Future}(x):=\lbrace y \mid x \leq y\rbrace$.

*For any $x$, the intersection of ${\rm Past}(x)$ with any antichain is finite. Similarily for ${\rm Future}(x)$.
Questions: 
   1. Is the orbit of any point $x$ by an automorphism $f$ of the poset finite?

   2. Is the group of automorphisms of this poset countable?

   3. As a polish group, is the group of automorphisms of the poset locally compact?

Thank you 
 A: If I've understood the hypotheses correctly, the covering relation gives a connected locally finite directed graph: starting from a point $x$, the maximal elements underneath it form a finite antichain, as do the elements covering it.  The automorphism group is therefore a second-countable totally disconnected locally compact group under the topology given by declaring the stabiliser of each point to be open.  The automorphism group as a whole will be countable if and only if every point stabiliser is finite and compact if and only if every orbit of the group is finite.
There's no reason for orbits of even individual automorphisms to be finite: for instance, your poset could be $\mathbb{Z}$ and the automorphism could be $x \mapsto x+1$.
Edit: Here's a simple example of how the automorphism group can be large.  Start with $\mathbb{Z}$, and replace every odd number $n$ with $|n|$ incomparable copies of itself.  The automorphism group is $H \rtimes \mathbb{Z}/2\mathbb{Z}$, where $H$ is a Cartesian product of symmetric groups of unbounded degree.  In particular the automorphism group is uncountable, and contains a copy of every countably-based profinite group.
A: Claim:  The poset is a chain.
Proof: Consider an unrelated pair a, b in the poset.  Let A = {a}.  Then a \in Past(b) by condition 1, which leads to a contradiction.
I imagine that your definitions are unclear.  Please carefully define 'covered by', 'maximal chain', poset?, etc.  
