If $X$ and $Y$ are equipotent then the bijections set of $X$ into $X$ is equipotent to the bijections set of $Y$ into $Y$.

Question

If $f$ is a bijection from a set $X$ to a set $Y$ then it is not complicated to prove that the equality $$ \tag{1}\label{eq:1}\big[\Phi_f(h)\big](y):=f\Big(h\big(f^{-1}(y)\big)\Big) $$ with $y\in Y$ defines a bijection from $X^X$ to $Y^Y$: in fact, if $h$ is in $X^X$ then for any $x$ in $X$ the equality $$ \big[(\Phi_{f^{-1}}\circ\Phi_f)(h)\big](x)=f^{-1}\Big(\big[\Phi_f(h)\big]\big(f(x)\big)\Big)=f^{-1}\biggl(f\Big(h\big(f^{-1}(f(x)\big)\Big)\biggl)=h(x) $$ holds and if $h$ is in $Y^Y$ then the equality $$ \big[(\Phi_f\circ\Phi_{f^{-1}})(h)\big](y)=f\Big(\big[\Phi_{f^{-1}}(h)\big]\big(f^{-1}(y)\big)\Big)=f\biggl(f^{-1}\Big(h\big(f(f^{-1}(y)\big)\Big)\biggl)=h(y) $$ holds; moreover, if $h$ in $X^X$ is a injection then the equality $$ \big[\Phi_f(h)\big](y_1)=\big[\Phi_f(h)\big](y_2) $$ with $y_1,y_2\in Y$ implies the equality $$ f\Big(h\big(f^{-1}(y_1)\big)\Big)=f\Big(h\big(f^{-1}(y_2)\big)\Big) $$ so that by inijectivity of $f$ and $h$ we conclude that $y_1$ and $y_2$ are equal and thus $\Phi_f(h)$ is an injection and thus by the equality $$ \Phi_{f^-1}=(\Phi_f)^{-1} $$ we conclude that the set of injection of $X$ into $X$ is equipotent to the set of injection of $Y$ into $Y$; moreover if $h$ in $X^X$ is a surjection then given $x$ in $h^{-1}[f^{-1}(y)]$ for any $y$ in $Y$ then the equality $$ \big[\Phi_f(h)\big]\big(f(x)\big)=f\Big(h\big(f^{-1}(f(x))\big)\Big)=f\big(h(x)\big)=f\big(f^{-1}(y)\big)=y $$ holds and so $\Phi_f(h)$ is a surjection and thus by the equality $$ \Phi_{f^{-1}}=(\Phi_f)^{-1} $$ we conclude that the set of surjections of $X$ onto $X$ is equipotent to the set of surjections of $X$ onto $Y$. So we conclude that if $h$ in $X^X$ is a bijection then $\Phi_f(h)$ is bijection and so actually $\Phi_f$ defines a bijection from the set of bijective functions $\mathcal B(X)$ of $X$ onto the set of bijective functions $\mathcal B(Y)$ of $Y$. Now if $X$ and $Y$ are equipotent then $X^X$ and $Y^Y$ are surely equipotent to $\kappa^\kappa$ where $\kappa$ is the cardinal of $X$ and $Y$ so that by transitivity $X^X$ and $Y^Y$ are equipotent and thus it is not necessary use $\Phi_f$ to prove that these set are equipotent: however is possible to prove the equipotency of $\mathcal B(X)$ and $\mathcal B(Y)$ without using $\Phi_f$? what happes then for the set of injections and of surjections? so could someone say something about?