 Solution

Put $|B|=m$. There are $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)m!=(m+1)!m/2$ surjections and $(m+1)!$ injections. For $m\ge 3$ the domain is larger, so $\Psi$ cannot be injective, regardless of the choices. For $m=1$ there is only one surjection, so it is injective. For $m=2$ both sets have size $6$; injectivity depends on the representative choices and, if achieved, is equivalent to bijectivity.