Whether a total order set of size $n$ has the fewest endomorphisms among posets of size $n$ 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?
 A: 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.


A: For posets with less than five elements, the total order is the unique poset with one unique endomorphism, but on 5 elements,
there are in total 3 posets with only one endomorphism:
the poset with edges
$$
12, 13, 24, 35, 45 \text{ and }
13, 14, 24, 25, 35
$$
Old unrelated answer:
Won't the unique poset $P$ on $n$ elements with no relations whatsoever 
admit the maximal number of endomorphisms?
Namely, any $P \to P$ is an endomorphism,
since any pair of elements are unrelated, and they are also mapped to a pair of unrelated elements. Or, if you do not allow two different elements with no relation to be mapped to the same element, then you get the $n!$ permutations.
