Conjugation of Bijection between Symmetric Groups is Isomorphism

Theorem
Let $A$ and $B$ be sets

Let $f$ be a bijection from $E$ to $F$.

Let $S_A$ and $S_B$ denote the set of all permutations on $A$ and $B$ respectively.

Let $\Phi: S_A \to S_B$ be the mapping defined as:
 * $\forall u \in S_A: \map \Phi u = f \circ u \circ f^{-1}$

where $\circ$ denotes composition of mappings.

Then $\Phi$ is an isomorphism from $S_A$ to $S_B$.

Proof
We have that $\struct {S_A, \circ}$ and $\struct {S_B, \circ}$ are the symmetric group on $S_A$ and $S_B$ respectively.

Hence we are about to prove that $\Phi$ is actually a group isomorphism.

Because $f$ is a bijection it follows from Inverse of Bijection is Bijection that $f^{-1}$ is also a bijection.

From Composite of Bijections is Bijection, it follows that $f \circ u \circ f^{-1}$ is also a bijection.

As $f \circ u \circ f^{-1}$ is from $S_B$ to $S_B$, it follows by definition that $f \circ u \circ f^{-1}$ is in fact a permutation on $B$.

Hence $\Phi$ maps a permutation on $A$ to a permutation on $B$, as stated by the question.

Let $u$ and $v$ be arbitrary permutations on $A$.

Then:

This demonstrates that $\Phi$ is a (group) homomorphism.

Let $u, v \in S_A$ such that $u = v$.

Then:

So we have:
 * $\map \Phi u = \map \Phi v \implies u = v$

and by definition $\Phi$ is injective.

Let $w \in S_B$.

Let $g: S_A \to S_B$ be defined as:
 * $g := f^{-1} \circ w \circ f$

Then from Inverse of Bijection is Bijection and Composite of Bijections is Bijection as above:
 * $g$ is a bijection from $S_A$ to $S_B$.

Thus we have:

Thus $\forall w \in S_B: \exists g \in S_A: \map \Phi g = w$

That is: $\Phi$ surjective.

Thus $\Phi$ has been shown to be a bijective (group) homomorphism.

Hence the result.