Sets of Permutations of Equivalent Sets are Equivalent

Theorem
Let $A$ and $B$ be sets such that:
 * $A \sim B$

where $\sim$ denotes set equivalence.

Let $\map \Gamma A$ denote the set of permutations on $A$. Then:
 * $\map \Gamma A \sim \map \Gamma B$

Proof
By definition of set equivalence, let $f: A \to B$ be a bijection.

Define $\Phi : \map \Gamma A \to \map \Gamma B$ by:
 * $\map \Phi \gamma := f \circ \gamma \circ f^{-1}$

By definition of permutation, each $\gamma \in \map \Gamma A$ is a bijection.

By Composite of Bijections is Bijection, each $f \circ \gamma$ is a bijection.

By Inverse of Bijection is Bijection, the inverse mapping $f^{-1}$ is a bijection.

Then by Composite of Bijections is Bijection, each $\map \Phi \gamma = f \circ \gamma \circ f^{-1}$ is a bijection.

Thus each $\map \Phi \gamma$ is a permutation of $B$, so $\Phi$ is well-defined.

Define $\Psi : \map \Gamma B \to \map \Gamma A$ by:
 * $\map \Psi \delta := f^{-1} \circ \delta \circ f$

$\Psi$ is a well-defined mapping by the same reasoning as for $\Phi$.

Then for each $\gamma \in \map \Gamma A$:

Thus $\Psi \circ \Phi = I_{\map \Gamma A}$.

Similarly, for each $\delta \in \map \Gamma B$:

Thus $\Phi \circ \Psi = I_{\map \Gamma B}$.

By Bijection iff Left and Right Inverse, both $\Phi$ and $\Psi$ are bijections.

In particular:
 * $\map \Gamma A \sim \map \Gamma B$