Condition for Isomorphism between Structures Induced by Permutations

Theorem
Let $S$ be a set.

Let $\oplus$ and $\otimes$ be closed operations on $S$ such that both $\oplus$ and $\otimes$ have the same identity.

Let $\sigma$ and $\tau$ be permutations on $S$.

Let $\oplus_\sigma$ and $\otimes_\tau$ be the operations on $S$ induced on $\oplus$ by $\sigma$ and on $\otimes$ by $\tau$ respectively:
 * $\forall x, y \in S: x \oplus_\sigma y := \map \sigma {x \oplus y}$
 * $\forall x, y \in S: x \otimes_\tau y := \map \tau {x \otimes y}$

Let $f: S \to S$ be a mapping.

Then:
 * $f$ is an isomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$


 * $f$ is an isomorphism from $\struct {S, \oplus}$ to $\struct {S, \otimes}$ satisfying the condition:
 * $f \circ \sigma = \tau \circ f$
 * where $\circ$ denotes composition of mappings.
 * where $\circ$ denotes composition of mappings.

Proof
Recall that:
 * an isomorphism is a bijection which is a homomorphism


 * a permutation is a bijection from a set to itself.

Hence on both sides of the double implication:
 * $f$ is a permutation on $S$


 * both $f \circ \sigma$ and $\tau \circ f$ are permutations on $S$.

So bijectivity of all relevant mappings can be taken for granted throughout the following.

Necessary Condition
Let $f$ be an isomorphism from $\struct {S, \oplus}$ to $\struct {S, \otimes}$ satisfying the condition:
 * $f \circ \sigma = \tau \circ f$

We have:

demonstrating that $f$ is a homomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$.

As $f$ is a bijection, it follows by definition that $f$ is an isomorphism from $\struct {S, \oplus_\sigma}$ to $\struct {S, \otimes_\tau}$.

Sufficient Condition
Let $f: \struct {S, \oplus_\sigma} \to \struct {S, \otimes_\tau}$ be an isomorphism.

Let $e \in S$ be the identity for both $\oplus$ and $\otimes$,.

We have:

Thus $f$ is a homomorphism, and thus an isomorphism, from $\struct {S, \oplus}$ to $\struct {S, \otimes}$.

Now we have:

In particular, this holds for $y = e$, so:

Hence by Equality of Mappings:
 * $\tau \circ f = f \circ \sigma$