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The first (and hardest) step is to factor $n$; the easiest way to do this (given $e$ and $d$) is with this randomized procedure: Select a random value $z$ from the range $(2, n-2)$ Compute the value $\lambda = (ed-1)/2^k$, where $k$ is that integer that makes $\lambda$ an odd integer. Compute $t = z^\lambda \bmod n$. If $t = 1$ or $t = n-1$, we fail on ...
Not quite, if all possible permutations are allowed. There are $8! = 40320$ permutations over 3 bits; 15 bits of key allows you to specify $2^{15} = 32768$ of them; hence any mapping of 15 bits will necessarily $40320-32768 = 7552$ of permutations unexpressable. It is doable if you don't allow every single permutation (e.g. allow only even permutations), ...