TITLE: Whether a total order set of size $n$ has the fewest endomorphisms among posets of size $n$
QUESTION [10 upvotes]: A function $f: P \to P$ is an endomorphism iff for any $x \le y$ in the poset $P$ , $f(x) \le f(y)$. So among posets of size $n$, whether the total order set $[n]$ (with the usual ordering) has the fewest endomorphisms?

REPLY [11 votes]: No. The zig-zag poset on 4 elements has only 31 endomorphism, whereas the total order has 35 endomorphisms.
I added the number of automorphisms and endomorphisms of a poset to http://www.findstat.org, should be visible shortly.
Update:
To make things a little bit clearer, note that adding more covering relations may increase the number of endomorphisms.  Consider the two posets below, then the map sending $[0,1,2,3]$ to $[0,0,0,2]$ is not a poset endomorphism of the fence, but it is a poset endomorphism of the second poset.