Let $X$ and $Y$ be two ordered sets, each of cardinality $6$. I need an example of two such sets which are not order isomorphic.
I tried with $(X,\le_X)=(\{\pm1,\pm2,\pm3\},\le_X)$ and $(Y,\le_Y)=\{\pm1,\pm4,\pm9\},\le_Y)$ and defined $f:X\to Y$ by $f(x)=x^2$ for all $x\in X$ and $\le$ is 'less than or equal to'.
Since $f$ is not surjective, so not bijective and consequently not order-isomorphic.
Is my explanation ok? If not, where did I make mistake? And if my answer is correct, is there any more convincing example other than this?
There are $318$ non-isomorphic posets with $6$ elements.
To give just two of them, consider the six-element chain and the six-element anti-chain:
take the set $X = \{ 1, 2, 3, 4, 5, 6 \}$ and define two order relations:
The first is the equality, that is two elements are related iff they are equal (it's the same element). It's easy to check this is a partial order relation (just check the defining properties). Its Hasse diagram is just a set of six nodes without any connections between them.
The second is, for example, the order given by restricting to this set the usual order on the natural numbers: $x \leq y$ iff $y - x$ is non-negative. The Hasse diagram of this poset is given by a set of six nodes starting from node $1$, and then node $2$ above it, with the corresponding line uniting them; then node $3$, ... and so on until node $6$ (five lines uniting them).