co.combinatorics - Is there a poset with 0 with countable automorphism group?

It seems unlikely (once you assume d.c.c.). Define the height of an element $x$ in $P$ to be the length of the shortest unrefinable chain from $x$ to $0$.

Let $P_n$ denote the elements of $P$ whose height is at most $n$. Since each element has a finite number of covers, the number of elements in $P_n$ is finite.

By d.c.c., every element of $P$ is in some $P_n$.

Let $G$ denote the automorphisms of $P$ and let $G_n$ denote the automorphisms of $P_n$. $G$ is the inverse limit of the system $G_n$. Let $H_n$ denote the image of $G$ inside $G_n$. (Note that this might not be all of $G_n$, since there could be automorphisms of $P_n$ that don't extend to $P$.) $G$ is also the inverse limit of the system $H_n$.

If the system $H_n$ stabilizes, then $G$ is finite. On the other hand, if $H_n$ doesn't stabilize, then the cardinality of $G$ is an infinite product, i.e. uncountable.