All possible total orderings of a finite set are isomorphic. What are some other examples of this phenomenon?

Question

All possible total orderings of a finite set are isomorphic. I find these kinds of results remarkable. Here's a few more.

Assume that $S$ is a finite set. Then:

1. All possible field structures on $S$ are isomorphic (and there exists a field structure on $S$ iff the cardinality of $S$ is $p^n$ for some prime $p$ and positive integer $n$.)
2. All possible Boolean algebra structures on $S$ are isomorphic (and there exists a Boolean algebra structure on $S$ iff the cardinality of $S$ is $2^n$ for some natural number $n$.)
3. All possible cyclic group structures on $S$ are isomorphic.
4. If the cardinality of $S$ is prime, then all possible ways of making $S$ into a group yield isomorphic groups.
5. All possible ways of making $S$ into a cycle (in the sense of graph theory) are isomorphic.

Question. Not necessarily assuming that $S$ is a set (e.g. it can be an abelian group, or a ring, or whatever) what are some other examples of this phenomenon?

We can make things a little more precise using the language of categories.

Given a functor $U : \mathbf{S} \leftarrow \mathbf{C}$ and an object $S$ of $\mathbf{S}$, define that:

• $U^{-1}(S)$ is the full subcategory of $\mathbf{C}$ consisting of precisely those objects whose image under $U$ is $S$.
• $U^{-1}(\mathrm{id}_S)$ is the wide subcategory of $U^{-1}(S)$ consisting of precisely those morphisms of $U^{-1}(S)$ whose image under $U$ is $\mathrm{id}_S$.

We're interested in results of the form: the category $U^{-1}(\mathrm{id}_S)$ has multiple non-isomorphic objects; nonetheless, all objects of $U^{-1}(S)$ are isomorphic.

Answer 1

All order 3 Latin squares on the symbols $A$, $B$, $C$ (there are 12 of them) are isotopic (that is, you can get from any one to any other by shuffling rows and/or columns and/or symbols).

Answer 2

For every field $F$ and every nonnegative integer $n$, all $n$-dimensional $F$-vector spaces are isomorphic.

Answer 3

All simple transcendental extensions of a given field are isomorphic.

Answer 4

In some sense, this is a nice generalization of some of your facts:

If $T$ is some complete theory, and $M \models T$, $|M|< \omega$, then if $N \models T$, we have that $M \cong N$.

Example: Let $L= \{<\}$, and let $M$ be a finite total order. Let $N \models Th_L(M)$. We show that $M \cong N$.

Recall that we can say that $M$ has exactly $m$ elements in FOL $\implies$ $N$ has exactly $m$ elements. Since $M$ is finite, $M$ has a first element ($(\exists x)(\forall y)( x \leq y))$, call it $0_m$ $\implies$ N has a first element, call it $0_n$. Since $M$ is finite, it's ordering is discrete (which can be written out in $FOL$) and sos every element has a direct successor $\implies$ every element of $N$ has a direct successor.

Now construct the function $f_0(0_m)= f_0(0_n)$, $f_k(S^k(0_m))= f_k(S^k(0_m))$ and let $f = \bigcup_{i=1}^{m-1}f_i$. One now just has to check that $f$ is an isomorphism.