Does every permutation (autojection) on a magma (binary algebra) induce another (not necessarily distinct) magma?

Question

Define a permutation or autojection A as a function which bijects from a set X to itself.

Define a binary algebra or magma as a set with a binary operation on it.

Let un-subscripted Hindu-Arabic numerals to denote binary operations.

Conjecture: For every binary algera $N_{1}$=(N, 1) on a set N with with n elements, where n belongs to {2, 3, ...} (equivalently, n equals any natural number greater than 1), there exists an algebra $N_{2}$=(N, 2) (not necessarily distinct from $N_{1}$) such that if A indicates any autojection on N, then A qualifies as an automorphism between $N_{1}$ and some $N_{2}$ (not between every $N_{1}$ and every $N_{2}$).

Question 1: Is this conjecture correct?

Question 2: If correct, how does one prove this conjecture?

This seems true intuitively to me, and it comes as true by definition that A indicates an autojection. So, it would seem that only the homomorphic equation xy1A=xAyA2 would need verified. But, how does one do verify it here, or ensure that the homomorphic equation holds with A as an autojection?

Answer 1

No, the conjecture is not correct. Let $N=\{a,b\}$, and define the operation $\star\,\,:N\times N\rightarrow N$ to be the constant function to $a$. Suppose $\bullet\,\,:N\times N\rightarrow N$ is any other operation on $N$.

If the image of $\bullet$ is all of $N$, then there cannot be any isomorphisms from $(N,\star)$ to $(N,\bullet)$.

If the image of $\bullet$ is $\{a\}$ (i.e. $\bullet=\star$), the autojection $p:N\rightarrow N$ with $p(a)=b$ and $p(b)=a$ is not an isomorphism from $(N,\star)$ to $(N,\bullet)$, because $$p(a\star a)=p(a)=b\neq a= p(a)\star p(a)=p(a)\bullet p(a).$$

If the image of $\bullet$ is $\{b\}$, then the identity autojection is not an isomorphism from $(N,\star)$ to $(N,\bullet)$ for a similar reason.

Answer 2

What is true is that given any autojection there is a binary algebra so that the autojection is an automorphism of binary algebras. You simply define the operation so that it works with the autojection. For instance, suppose you have the binary algebra $(N, \star)$, and an autojection $A: N \rightarrow N$. Then define a new operation $\bullet$ by $a \bullet b = A^{-1}(A(a)\star A(b))$ for all $a, b \in N$.

Answer 3

Given any two sets $A$ and $B$, and a bijection $f\colon A\to B$, any $\alpha$-ary operation $\tau$ (for $\alpha\in\mathrm{Ord}$) yields an $\alpha$-ary operation $\tau'$ on $B$ defined by $$\tau'\left(\{b_i\}_{i\in\alpha}\right) = f\left(\tau\left(\{g(b_i)\}_{i\in\alpha}\right)\right),$$ where $g\colon B\to A$ is the set-theoretic inverse of $f$. Under this definition, $f\colon (A,\tau)\to (B,\tau')$ respects the operation; i.e., $$f\left(\tau(\{a_i\}_{i\in\alpha})\right) = \tau'\left(\{f(a_i)\}_{i\in \alpha}\right).$$

(This includes binary, ternary, $n$-ary, $\omega$-ary, unary, nullary, etc. operations; and of course, $B$ can be $A$).

This easily extends to any family (or even proper class) of operations defined on $A$, so that given a bijection $f\colon A\to B$ and a class $\Omega$ of operations on $A$, we can define on $B$ a family of operations of the same types as $\Omega$ that make $f$ into an $\Omega$-algebra isomorphism.

And stripped of all the formalism and all the neologisms, what this says is basically that if you change the names of all the elements, but keep the operations the same, you get an isomorphic structure. E.g., addition in the natural numbers in English is isomorphic to addition in the natural numbers in Spanish (just be careful in the translation; remember that $\text{one billion}\neq\text{un billón}$, just like $\text{library}\neq\text{librería}$.)