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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 $H \rtimes D_\infty$, where $D_\infty$ is the infinite dihedral group and $H$ is a Cartesian product of symmetric groups of unbounded degree. In particular the automorphism group is neither compact nor countable, and contains a copy of every countably-based profinite group.

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.

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 $H \rtimes D_\infty$, where $D_\infty$ is the infinite dihedral group and $H$ is a Cartesian product of symmetric groups of unbounded degree. This group is neither compact nor countable, and contains a copy of every countably-based profinite group.

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 $H \rtimes D_\infty$, where $D_\infty$ is the infinite dihedral group and $H$ is a Cartesian product of symmetric groups of unbounded degree. In particular the automorphism group is neither compact nor countable, and contains a copy of every countably-based profinite group.

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. I can't think of a nice example right now, but I'm sure there are examples where the point stabilisers are 'large' in some sense, such as containing a copy of every countably-based profinite group.

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 $x \mapsto x+1$.

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 $H \rtimes D_\infty$, where $D_\infty$ is the infinite dihedral group and $H$ is a Cartesian product of symmetric groups of unbounded degree. This group is neither compact nor countable, and contains a copy of every countably-based profinite group.