Models of a certain (weird) equational theory Consider the following (single-sorted) equational/algebraic theory with one binary operation symbol $\ast$ whose axioms are as follows:
$$(x \ast x) \ast (x \ast x) = x$$
$$(x \ast y) \ast (x \ast y) = (x \ast x) \ast (y \ast y).$$
I am interested in models of this theory where $\ast$ is NOT idempotent, i.e. where $x \ast x = x$ is not true for every $x$ in the model. So far, I have come up with the following toy model of this theory, where $\ast$ is not idempotent: the carrier is $\{0, 1\}$, and the binary operation $\ast$ is defined as follows: $$0 \ast 0 = 1,$$ $$1 \ast 1 = 0,$$ $$0 \ast 1 = 0,$$ $$1 \ast 0 = 1.$$
My question is, are there any more 'natural' models of this theory where $\ast$ is NOT idempotent, i.e. are there any non-idempotent binary operations satisfying the above axioms that have been previously studied in mathematics?
 A: There is such a theory that was actually going to be the topic of my dissertation. Let $R$ be a root system, let $x,y$ be roots, and let $s_x$ be the reflection in the hyperplane normal to $x$. Then we may define
$$x*y = s_x(y)$$
Then there does not exist any $x$ for which $x*x=x$. In fact, for all $x$ we have
$$x*x = -x$$
We therefore have
$$(x*x)*(x*x) = - (-x) = x$$
Note we have 
$$x*(-y)=-(x*y)$$
and
$$(x*x)*y = x*y$$
We also have
$$(x*y)*(x*y)=-(x*y)$$
and
$$(x*x)*(y*y) = x*(-y) = -(x*y)$$
so
$$(x*x)*(y*y)=(x*y)*(x*y)$$
The two axioms necessary to prove these things are


*

*$x*(x*y)=y$

*$x*(y*z)=(x*y)*(x*z)$
There is a third axiom that ensures that the resulting algebra is a root system, but it is a bit more of a pain to state.
A: This answer expands on a comment of Captain Lama; if they post an answer of their own I'll delete this one, and I've made it community-wiki so I don't get reputation for their work.
Note that the second condition is an immediate consequence of associativity and commutativity. So any commutative semigroup satisfying $x^4=x$ will satisfy your theory - for example, the group $\mathbb{Z}/3\mathbb{Z}$.
(Of course, there are structures satisfying your theory which are not commutative semigroups, but commutative semigroups are relatively simple things to think about.)
A: Let me describe how to produce typical
models of this equational theory using a
different but equivalent language.
First, number the two given axioms:
Axiom $(1)$. $(x*x)*(x*x)=x$ 
Axiom $(2)$. $(x*y)*(x*y)=(x*x)*(y*y)$ 
Let $\sigma(x)=x*x$ be the squaring map with respect to $*$,
and let $x\odot y=\sigma(x*y)$. 
Axiom (1) asserts exactly that $\sigma$ is a permutation of
exponent $2$, while Axiom (2) asserts exactly that $\sigma$
commutes with $*$. Since $\sigma$ also commutes with itself,
it will then commute with $\odot$, which is a composition
of $\sigma$ and $*$. Since $x\odot y$ is defined to be
$\sigma(x*y)$, and $\sigma$ has exponent $2$, we can recover
$*$ from $\sigma$ and $\odot$ by $x*y=\sigma(\sigma(x*y))=\sigma(x\odot y)$.
Altogether, this shows that we can convert between the
$*$-language and the $\odot,\sigma$-language using
these definitions:

* $\sigma(x):=x*x$.

* $x\odot y:=\sigma(x*y)$.

* $x*y:=\sigma(x\odot y)$.
Now, in order to translate theories, we observe that
an algebra $\langle A; *\rangle$ in the language
$\{*\}$ satisfies Axioms (1) and (2)
iff the corresponding algebra
$\langle A; \odot, \sigma\rangle$ in the language
$\{\sigma,\odot\}$ satisfies
Axiom $(1)'$. the binary operation of $\langle A; \odot\rangle$ is idempotent, and 
Axiom $(2)'$. $\sigma$ is an exponent-2 automorphism of
$\langle A; \odot\rangle$.
That is, up to a change of language,
a model of the original axioms is simply an
idempotent binary algebra equipped with
an exponent-$2$ automorphism. 
Examples.

* It is not hard to characterize the
examples where $\sigma$ is trivial
(i.e., $\sigma$ is the identity function).
Any such algebra is obtained from an idempotent binary algebra
$\langle A; \odot\rangle$ by setting $x*y=x\odot y$.

* It is not hard to characterize 
the examples where $\odot$ is trivial
(i.e., $\odot$ is one of the projections
$x\odot y = x$ for all $x, y$ or
$x\odot y = y$ for all $x, y$).
In this case, for any set $A$ let $\sigma: A\to A$
be any permutation of exponent $2$ ($\sigma^2(x)=x$).
Then $x*y:=\sigma(x)$ or $x*y:=\sigma(y)$ are both
operations on $A$ satisfying Axioms (1) and (2).
The example in the question statement
is of this type.

* Let $\mathbb A = \langle A; \odot\rangle$ be any idempotent
binary algebra. Let $\mathbb B = \mathbb A\times \mathbb A$.
Let $\sigma: \mathbb B\to \mathbb B: (b,c)\mapsto (c,b)$
be the automorphism of switching coordinates.
A change of language converts
$\langle B; \odot, \sigma\rangle$ into a model of
Axioms (1) and (2).

* Let $M$ be an $R$-module. Suppose that $r,s\in R$ commute with each other and $s^2=1$. Then $x\odot y:=rx+(1-r)y$ is idempotent and $\sigma(x)=sx$ is an exponent 2 automorphism of $\langle M; \odot\rangle$, so if we equip $M$ with only the operation $x*y=\sigma(x\odot y) = srx+s(1-r)y$, then $\langle M; *\rangle$  will satisfy Axioms (1) and (2).
