Existence of a homogeneous cyclic order for arbitrary infinite cardinals It seems striking that the cardinalities of $\aleph_0$ and $\mathfrak c = 2^{\aleph_0}$ each admit what I will call a "homogeneous cyclic order", via the examples of $ℚ/ℤ$ and $ℝ/ℤ$. By which I mean a cyclic order (as defined in https://ncatlab.org/nlab/show/cyclic+order) such that for any two elements $x, y$ of the cardinal, there is a bijection of the cardinal to itself taking $x$ to $y$ and preserving the cyclic order.
In ZFC, is there any reason to believe that either a) all infinite cardinals admit a homogeneous cyclic order, or b) there exists an infinite cardinal admitting no homogeneous cyclic order?
 A: Adjoin to the theory of cyclic orders (as on the ncatlab page linked in the question) a 3-place function $f$ and axioms saying that, for each fixed $x$ and $y$, the function $z\mapsto f(x,y,z)$ is a bijection that respects $R$ (i.e., and automorphism of the underlying $R$-strtucture) and sends $x$ to $y$. The resulting first-order theory, in the language $\{R,f\}$, has infinite models (as noted in the question), so the upward and downward Löwenheim-Skolem theorems imply that it has models of all infinite cardinalities.
A: Every totally ordered abelian group is automatically homogeneous as a total order, and every totally ordered set that is homogeneous as a total order is automatically homogeneous as a cyclic order. Therefore, every totally ordered abelian group is automatically homogeneous as a cyclic order.
Let $\lambda$ be an infinite cardinal. Let $G_{\alpha}$ be a totally ordered abelian group for $\alpha<\lambda$. Then the abelian group coproduct $\oplus_{\alpha<\lambda}G_{\alpha}$ is an abelian group which is totally ordered by letting
$(x_{\alpha})_{\alpha<\lambda}<(y_{\alpha})_{\alpha<\lambda}$ precisely when there is some ordinal $\gamma$ with $x_\gamma<y_\gamma$ but where $x_\beta=y_\beta$ for $\beta<\gamma$.
Since $|\oplus_{\alpha<\lambda}G_\alpha|=\sum_{\alpha<\lambda}|G_\alpha|$, we can make $\oplus_{\alpha<\lambda}G_\alpha$ have any infinite cardinality.
A: While Andreas's answer is a very simple construction let me point out that one can get even more.
Theorem: for every cardinal $\lambda$ there exists a ultra-homogeneous circularly ordered set $(X,R)$ with $|X|=\lambda$ (ultra-homogeneous here means that any isomorphism between finite substructures of $X$ extends to an automorphism of $X$).
This has been noted independentently by Pestov and Truss, but neither of them published it. It was published, with Pestov's proof, by Glasner and Megrelishvili in Circular orders, ultra-homogeneous order structures and
their automorphism groups, the construction below is due to Pestov and taken from the paper just mentioned.
Proof: Let $k$ be a linearly ordered field of size $\lambda$ (to construct one start with $\Bbb Q$ and do a tower of transcendental extensions, for $\alpha<\lambda$. At successor stages $\Bbb Q_{\alpha+1}=\Bbb Q_\alpha(x_\alpha)$ ordered in the usual way that makes $x_\alpha$ a positive infinitesimal over $\Bbb Q_\alpha$, while at limit stages take the union of the previous extensions). Now for $x\in k$ let $|x|=\max\{-x,x\}$ and let $\mathrm{fin}(k)=\{x\in k\mid \exists n\in\Bbb Z(|x|<n)\}$ (note that $k$ has characteristic zero being ordered). Now we mimic the $\Bbb R/\Bbb Z$ construction by considering $\mathrm{fin}(k)/\Bbb Z$. We set $(a,b,c)\in R$ if and only if there are $a'\leq b'\leq c'$ with $a=a'+\Bbb Z$, $b=b'+\Bbb Z$ and $c=c'+\Bbb Z$.
It remains to check that $\mathrm{fin}(k)/\Bbb Z$ is ultra-homogeneous. Let $a_1,\ldots,a_n$ and $b_1,\ldots,b_n$ be two positively circularly ordered subsets of $\mathrm{fin}(k)/\Bbb Z$, meaning that whenever $[i,j,k]$ in the circularly ordered finite group $\Bbb Z_k$, one has $[a_i,a_j,a_k]$ and $[b_i,b_j,b_k]$ in $\mathrm{fin}(k)/\Bbb Z$. Wlog assume $a_1=b_1=0$. Identifying $\mathrm{fin}(k)/\Bbb Z$ with the interval $[0,1)$ in $k$ we get representatives $0=a_1'<a_2'<\ldots<a_n'<1$ and $0=b_1'<b_2'<\ldots<b_n'<1$. Now apply a piecewise linear transformation $[0,1)_k\to[0,1)_k$ mapping $a_i'$ to $b_i'$ for every $i$, and lift it to an automorphism of $\mathrm{fin}(k)/\Bbb Z$ mapping $a_i$ to $b_i$ for every $i$.
