Do almost-point-transitive algebras generate almost-point-transitive varieties? Say that an algebra $\mathfrak{A}$ (in the sense of universal algebra) is point-transitive iff for every $a,b\in\mathfrak{A}$ there is a $\pi\in Aut(\mathfrak{A})$ with $\pi(a)=b$. While genuinely point-transitive algebras are somewhat rare, many naturally-occurring algebras yield the same clone as a point-transitive algebra. Specifically, given an algebra $\mathfrak{A}$ let $\mathsf{Cl}(\mathfrak{A})$ be the smallest set of functions from some finite Cartesian power of $\mathfrak{A}$ to $\mathfrak{A}$ which contains all the constant functions and projection functions, all the primitive functions of $\mathfrak{A}$ itself, and is closed under composition. (EDIT: as Keith Kearnes states below, this is actually the polynomial clone of $\mathfrak{A}$.) Then:

*

*Each group $(G;*,{}^{-1},e)$ yields the same clone as its "torsor reduct" $(G;(a,b,c)\mapsto a*(b^{-1}*c))$, which is clearly point-transitive since each $x\mapsto y*x$ is an automorphism.


*A similar trick works with rings: each ring $(R;0,1,+,-,\times)$ yields the same clone as $(R; (a,b,c,d,e)\mapsto (a-b)(c-d)+e)$, and the latter is again point-transitive via maps of the form $x\mapsto x+y$. (This is basically due to Matt F.)
Say that an algebra $\mathfrak{A}$ is almost-point-transitive iff we have $\mathsf{Cl}(\mathfrak{A})=\mathsf{Cl}(\mathfrak{B})$ for some point-transitive algebra $\mathfrak{B}$ with the same underlying set. I'd like to just ask "Which algebras are almost-point-transitive?," but I don't see how to make that precise in the right way (e.g. to avoid "the almost-point-transitive ones" as an answer). Instead, the following seems like it might be more immediately approachable:

If $\mathfrak{A}$ is almost-point-transitive and $\mathfrak{B}\in\mathsf{HSP}(\mathfrak{A})$, must $\mathfrak{B}$ be almost-point-transitive as well?

I strongly suspect that the answer is negative but I don't see how to construct a counterexample.
 A: Let me distinguish between clone and polynomial clone.
The former is the smallest composition-closed
collection of operations on $A$
containing the primitive operations of $\mathbb A$ and the projections,
while the latter is the smallest composition-closed
collection of operations on $A$
containing the primitive operations of $\mathbb A$, the projections,
and the constant operations.
In fact, let me write $\mathbb A_A$ for the constant expansion of $\mathbb A$,
and then refer to the clone of $\mathbb A_A$ when I want to
talk about the polynomial clone of $\mathbb A$.

The $1$-unary
algebra $\mathbb A =\langle \mathbb Z; f(x)\rangle$,
where $f^{\mathbb A}(x)=x+1$,
provides a negative answer to the question.
This algebra has transitive automorphism group,
since all powers of $f^{\mathbb A}(x)$ are automorphisms.
The terms in this language in the variable
$x$ are $x, f(x), f^2(x), \ldots$, so each proper subvariety
of the variety of all $1$-unary algebras is
axiomatizable by some set of
identities of the form $f^m(x)\approx f^n(x)$,
$m\neq n$, or of the form $f^m(x)\approx f^n(y)$.
None of these hold in $\mathbb A$, so $\mathbb A$
generates the variety of all $1$-unary algebras.
Let $\mathbb B = \langle \mathbb Z; f(x)\rangle$
where $f^{\mathbb B}(x)=|x|$.
$\mathbb B$ is a $1$-unary algebra whose basic operation
has proper range.
The algebra $\mathbb B_B$ has the property
that its clone consists of essentially unary
operations only, it contains all the constant
operations on the domain, and it contains
a nonsurjective unary operation $f^{\mathbb B}(x)$.
Let $\mathbb C$ be any algebra defined on the same
set as $\mathbb B$ whose polynomial clone
agrees with that of $\mathbb B$.
Then $\mathbb C_C$ has the property
that its clone consists of essentially unary
operations only, it contains all the constant
operations on the domain, and it contains
a nonsurjective unary operation $f^{\mathbb B}(x)$.
Since the clone of the reduct $\mathbb C$
contains all the nonconstant operations
of the clone of $\mathbb C_C$,
the clone of $\mathbb C$
must contain $f^{\mathbb B}(x)$.
The range of $f^{\mathbb B}(x)$ is closed under
any automorphism of $\mathbb C$, so the automorphism
group of $\mathbb C$ is not transitive.
Altogether this shows that (i) $\mathbb A$
has transitive automorphism group (so $\mathbb A$ is point-transitive), but
(ii) the variety generated by $\mathbb A$
contains an algebra $\mathbb B$
that is not almost-point-transitive.
