Are well-orderings and converse well-orderings the only partial orderings which are uniquely isomorphic? If a partial order $(P, \leq)$ is a well-order or a converse well-order, there can be at most one isomorphism between it and any other partial order. Are well-orders and converse well-orders the only partial orders with that property?
 A: No.
The easy way to construct an infinite family of counterexamples is to take non-isomorphic well-orders $P$ and $Q$ and construct $S = P \dot\cup Q$ the partial order whose universe is the disjoint union between $P$ and $Q$ and the order relation is the inherited order from $P$ and $Q$ on any pairs from the original sets and for pairs in between, the relation never holds.
Now since $P$ and $Q$ are not isomorphic well-orders, any isomorphism of a partial order into $P\dot\cup Q$ will need to uniquely map one part to $P$ and one part to $Q$ as a biproduct of the theorem you mention. This method can then create a lot of new partial orders.
For an example of something which is not inherited from well-orders you can use the divisibility order on $\{1,2,3,4\}$ (i.e. $1 <2<4$ and $1<3$ and nothing else holds) just like Arthur mentions in his comment.
A: The answer is no. Let me reframe this question as a question about rigidity, and then send you to this wonderful answer by Andrés Caicedo to a question I had a few years ago regarding rigidity and well-ordering.
If $(A,<)$ is a rigid partial-ordering, and $(B,<')$ is isomorphic to $A$, then the isomorphism has to be unique. This is because given $f,g\colon A\to B$ which are isomorphisms, $f^{-1}\circ g$ is an automorphism of $A$ and is therefore the identity.
So we can change your question and ask, is it true that only well-ordered sets and co-well-ordered sets have this property?
Well, even if you require linearity the answer is still no, since $\omega+\omega^*$ (which is the order-type of $\{\frac1k\mid k\in\Bbb Z\setminus\{0\}\}$ as a set of reals) has no automorphisms, and it is neither well-ordered nor co-well-ordered.
To find out a lot more, you should read the answer I linked to.
