Theorem: Conjugation of Bijection between Symmetric Groups is Isomorphism

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:
{{begin-eqn}}
{{eqn | l = \map \Phi u \circ \map \Phi v
      | r = \paren {f \circ u \circ f^{-1} } \circ \paren {f \circ v \circ f^{-1} }
      | c = Definition of $\Phi$
}}
{{eqn | r = f \circ u \circ \paren {f^{-1} \circ f} \circ v \circ f^{-1}
      | c = Composition of Mappings is Associative
}}
{{eqn | r = f \circ u \circ I_A \circ v \circ f^{-1}
      | c = Composite of Bijection with Inverse is Identity Mapping
}}
{{eqn | r = f \circ \paren {u \circ v} \circ f^{-1}
      | c = {{Defof|Identity Mapping}} and Composition of Mappings is Associative
}}
{{eqn | r = \map \Phi {u \circ v}
      | c = Definition of $\Phi$
}}
{{end-eqn}}
This demonstrates that $\Phi$ is a (group) homomorphism.
Let $u, v \in S_A$ such that $u = v$.
Then:
{{begin-eqn}}
{{eqn | l = \map \Phi u
      | r = \map \Phi v
      | c = 
}}
{{eqn | ll= \leadsto
      | l = f \circ u \circ f^{-1}
      | r = f \circ v \circ f^{-1}
      | c = Definition of $\Phi$
}}
{{eqn | ll= \leadsto
      | l = \paren {f^{-1} \circ f} \circ u \circ \paren {f^{-1} \circ f}
      | r = \paren {f^{-1} \circ f} \circ v \circ \paren {f^{-1} \circ f}
      | c = applying $f^{-1}$ and $f$ to either end, and Composition of Mappings is Associative
}}
{{eqn | ll= \leadsto
      | l = I_S \circ u \circ I_S
      | r = I_S \circ v \circ I_S
      | c = Composite of Bijection with Inverse is Identity Mapping
}}
{{eqn | ll= \leadsto
      | l = u
      | r = v
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map \Phi g
      | r = f \circ \paren {f^{-1} \circ w \circ f} \circ f^{-1}
      | c = Definitions of $\Phi$ and $g$
}}
{{eqn | r = \paren {f \circ f^{-1} } \circ w \circ \paren {f \circ f^{-1} }
      | c = Composition of Mappings is Associative
}}
{{eqn | r = I_T \circ w \circ I_T
      | c = Composite of Bijection with Inverse is Identity Mapping
}}
{{eqn | r = w
      | c = {{Defof|Identity Mapping}}
}}
{{end-eqn}}
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.
{{qed}}