For any integer $k\geq 2$, a $k$-regular linear set system is a set ${\cal E}\subseteq {\cal P}(\omega)$ such that $|e| = k$ for all $e\in {\cal E}$, and moreover, for all $a\neq b\in\omega$ there is exactly one $e\in {\cal E}$ such that $\{a,b\}\subseteq e$. ([There is indeed](https://mathoverflow.net/a/428639/8628) a $k$-regular linear set system on $\omega$ for any given integer $k\geq 2$.)


We say that a set system ${\cal A}\subseteq {\cal P}(\omega)$ is *vertex-transitive* if for all $a, b\in \omega$ there is a bijection $\varphi:\omega\to\omega$ such that


1. $\varphi(a) = b$, and
2. for all $X\in{\cal A}$ we have $\varphi(X) \in {\cal A}$ as well as $\varphi^{-1}(X)\in {\cal A}$.


**Question.** If $k\geq 2$ is an integer and ${\cal E}\subseteq{\cal P}(\omega)$ is a $k$-regular linear system, is ${\cal E}$ necessarily vertex-transitive?



Are $k$-regular linear set systems vertex-transitive?
**Introduction**


I'll provide plenty of non-transitive examples (*just be patient, please*).


Let me introduce a useful and more general notion of *selector system* instead of *linear set system*; then, I will specialize it to the *linear* ones. even in the special case, we can equivalently consider an arbitrary infinite countable set in place of $\omega$.


**Remark** We will see that my complete (2 n)-systems are the same as the OP's linear set systems.




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**Definition of the systems and extensions**


A selector system is an ordered triple $\ (X\ \cal B\,\ \sigma)\ $ (*please, do not edit my comma-free notation*), where $\ X\ $ is an arbitrary set, $\ \cal B\subseteq 2^X,\ $ and $\ \sigma:B\to2^X\ $ satisfies two axioms:


$$ \forall\_{B\in\cal B}\quad B\cap\sigma(B)\ =\ \emptyset $$
and
$$\forall\_{A\ B\in\cal B}\quad(A\subseteq B\cup\sigma(B)\,\ \Rightarrow
 \,\ A\cup\sigma(A)\ =\ B\cup\sigma(B)) $$


Selector system $\ (Y\ \cal D\,\ \phi)\ $ extends selector system
$\ (X\ \cal B\,\ \sigma)\ \Leftarrow:\Rightarrow $ $ X\subseteq Y,
\ \cal B\subseteq\cal D,\ \sigma=\phi|\cal B,\ $ and


$$ \forall\_{B\in \cal B}\,\forall\_{D\in\ \cal D}
 \quad (D\subseteq B\cup\sigma(B)\,\ \Rightarrow\,\ B\cup\sigma(B)
 \ =\ D\cup\phi(D)) $$




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(k n)-**systems and their *free style* purposeful extensions**


Let $\ 1<k<n\ $ be natural numbers. In this section, let selector system $\ (X\ \cal B\,\ \sigma)\ $ be such that
$\ \emptyset\ne\cal B\subseteq$ $\binom Xk,\ $
$\ |\sigma(B)|=n-k,\ $ and


$$ \binom{B\cup \sigma(B)}k\subseteq \cal B $$
for every $\ B\in\cal B.\ $ Then we can construct an extension
$\ (Y\ \cal D\,\ \phi)\ $ that is similar (as we will see soon), and that has also the following ***purposeful property***:
$$ \binom Xk\ \subseteq\ \cal D $$
**Let's do it:**


Let set $\ T\ $ has cardinality $\ |T|=n-k,\ $ and let $\ Y\ $ be the
following disjoint "$\cup$":


$$ Y\,\ :=\,\ X\ \cup\ \left(\binom Xk\setminus\cal B \right)\times T, $$


and
$$ \cal D\ :=\ \cal B\,\cup
 \bigcup\_{b\in\binom Xk\setminus\cal B}\binom{b\cup\phi(b)}k,$$


*(sometimes the MO software forces calligraphic $\ \cal X\ $ in place of simple $\ X$)*


where $\ \phi\ $ extends $\ \sigma\ $ as follows:


$$ \forall\_{b\in\ \binom X2\setminus\cal B}\quad \phi(b)\ :=
 \ \{b\}\times T $$
and


$$ \forall\_{b\in\ \binom Xk\setminus\cal B}\,
 \forall\_{c\in\binom{b\cup\phi(b)}k}\quad\phi(c):=(b\cup\phi(b))\setminus c
 $$


Otherwise, $\ \phi(B)=\sigma(B)\ $ for $\ b\in\cal B.$


Thus, if $\ X\ $ is countable then so is $\ Y.$


The above extension was done in the ***free style***.




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**Complete selector** (k n)-**systems**


Now, we get an entire class of infinite countable ***complete*** selector systems
$\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty) $ such that
$\ B\_\infty=\binom {X\_\infty}k,\ $ and
$$ \forall\_{b\in\binom {X\_\infty}k} |\sigma\_\infty(b)|=n-k $$


***Remark**:   In general, the complete systems don't have to have the triple index $\infty$ in their names.*


We obtain a system as this as an extension of arbitrary countable system
$$ (X\_0\ \cal B\_0\ \sigma\_0)\ :=\ (X\ \cal B\,\ \sigma), $$
where
$$ \binom Xk\setminus\cal B\ \ne\ \emptyset $$


Then we obtain $\ X\_{s+1}\ \cal B\_{s+1}\ \sigma\_{s+1}\ $ from
$\ (X\_s\ \cal B\_s\ \sigma\_s)\ $ the way we have obtained
$\ (Y\ \cal D\,\ \phi)\ $ from $\ (X\ \cal B\,\ \sigma)\ $ for every $\ s\in\mathbb N.$ Now we define:


$$ X\_\infty\ :=\ \bigcup\_{s=0}^\infty\, X\_s\qquad\text{and}
 \qquad\cal B\_\infty\ :=\ \bigcup\_{s=0}^\infty\,\cal B\_s, $$
and let $\ \sigma\_\infty\ $ be the common extension of all $\ \sigma\_s,\ $
onto $\ \cal B\_\infty.$


THe above complete extension was done in the ***complete free style***.




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**No certain intermediate subsystems**


Let $\ (X\_0\ \cal B\_0\ \sigma\_0)\ :=\ $ $(X\ \cal B\,\ \sigma)\ $ be an
(k n)-system, and let $\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty)\ $ be
the complete extension of $\ (X\ \cal B\,\ \sigma)\ $ done in the complete
free style.


We also assume the non-completeness condition:
$$ \binom Xk\setminus\cal B\ne\emptyset $$




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**Theorem**   There does not exists an intermediate **complete**
subsystems $\ (Y\ \cal D\,\ \phi\ ) $ such that
$$ \qquad 0\ <\ |Y\setminus X|\ <\ \infty\qquad\qquad\qquad\qquad (\#) $$




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**Proof**


By (#), there exists $\ s\in\mathbb N\ $ such that
$$ Y\setminus X\_s\ =\ \emptyset\ \ne\ Y\setminus X\_{s-1} $$
Let $\ x\in Y\setminus X\_{s-1}.\ $ Then, for arbitrary
$\ L\subseteq\binom{X\_{s-1}}{k-1}\ne\emptyset\ $ and $\ p:=\{x\}\cup L\ $
we have
$$ \emptyset\ \ne\ \sigma\_{s+1}(p)\ \subseteq\ X\_{s+1}\setminus X\_n $$
Thus, $\ p\notin\binom Yk.\quad$ **End of Proof**




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**Non-transitive examples**


Let (k n)-system $\ (X\_1\ \cal B\_{-1}\ \sigma\_1)\ $ be complete, i.e.
$$ B\_{-1}\ =\ \binom{X\_{-1}}k $$


The examples can be taken from finite geometries or trivial (finite or nor)
when $\ |X\_{-1}|=n\ $ hence necessarily $\ \cal B\_{-1}:=\binom{X\_{-1}}k$.


Let $\ e\ $ and $\ X\_0\ :=\ X\_0\cup\{e\}\ $ be such that
$\ e\in X\_0\setminus X\_{-1}.$
Then we get an induce non-complete (k n)-system
$\ (X\_0\ \cal B\_0\ \sigma\_0),\ $ where
$$ X\_0:=X\_{-1}\cup{e}\qquad\text{and}\qquad\cal B\_0=\cal B\_{-1}\qquad
 \text{and}\qquad\forall\_{B\in\cal B\_0}\ \sigma\_0(B):=\sigma\_{-1}(B) $$




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**Theorem**   In every such situation as above, if $\ X\_0|<\infty\ $
then


$\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty)\ $ is an infinite complete
(k n)-system that is not vertex transitive.




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**Proof**   Indeed, there is no authomorphism $\ X\_\infty\to X\_\infty $
can map any element of $\ X\_{-1}\ $ to any element of
$\ X\_\infty\setminus X\_0.$     **End of Proof**


