For question 1, the automorphism group needn't be countable. To see this, consider the poset consisting of infinitely many diamonds stacked on top of one another.

.
.
.
3
/ \
* *
\ /
2
/ \
* *
\ /
1
/ \
* *
\ /
0

This is locally finite, countable and connected in the sense you have requested. Your additional property is satisfied because every antichain (in the sense of pairwise incomparable elements) has size at most two. The automorphism group is uncountable, however, because we can swap the two intermediate points of any diamond independently, and so any infinite binary sequence determines a distinct manner of performing these swaps.

There is a natural topology on the automorphism group here, as in your other question, whose basic open sets are determined by a finite piece of the automorphism. Indeed, the group is essentially the same as Cantor space $2^{\mathbb{N}}$, since there is a one-to-one correspondence between the binary sequences and the automorphisms, just by considering the diamonds in which the swap is made or not. This space has a natural probability measure, treating the swaps as coin flips. Thus, the collection of automorphisms that make $n$ many prescribed swaps or non-swaps has measure $2^{-n}$. This measure interacts well with the topology, because they are both the standard concepts on Cantor space.

Finally, to address Gerhard's idea of building a poset satisfying the properties but having a countably infinite automorphism group, while having diamonds, let us simply join the diamonds horizontally:

* * * *
/ \ / \ / \ / \
... -2 -1 0 1 2 ...
\ / \ / \ / \ /
* * * *

This poset is locally finite, countable and connected, and it exhibits the extra property because every point has only finitely many successors and predecessors. Meanwhile, the automorphism group is countably infinite because it is precisely the infinite dihedral group: we have exactly the horizontal translations and reflections. (And to satisfy Gerhard, we have many diamonds!)

arbitrarycountable structure (in the model-theoretic sense, i.e., a set equipped with a bunch of finitary operations and relations), then $\mathrm{Aut}(M)$ carries a canonical topology making it a Polish group, namely the one inherited from the inclusion of $\mathrm{Aut}(M)$ in the product space $M^M$, where $M$ is given the discrete topology. – Emil Jeřábek Sep 19 '11 at 16:47