Everyday I'm Shuffling

Using the information provided, we can almost find $\sigma^2$ exactly, where $\sigma$ is the positional permutation function from $[1, 12] \rightarrow [1, 12]$ . The catch is the two 'N's:

$\sigma^2(1) = 7,\:\sigma^2(2) = 5,\:\sigma^2(3) = 12,\:\sigma^2(4) = 10$

$\sigma^2(5) = 9,\:\sigma^2(6) \in \{2, 4\},\:\sigma^2(7) = 11,\:\sigma^2(8) = 3$

$\sigma^2(9) \in \{2, 4\},\:\sigma^2(10) = 1,\:\sigma^2(11) = 8,\:\sigma^2(12) = 6$

If we postulate that $\sigma^2(6) = 2$ , we get the chain, for $\sigma^2$ :

$1 \rightarrow 7 \rightarrow 11 \rightarrow 8 \rightarrow 3 \rightarrow 12 \rightarrow 6 \rightarrow 2 \rightarrow 5 \rightarrow 9 \rightarrow 4 \rightarrow 10 \rightarrow 1$

However, this implies that $\sigma$ consists of a single cycle, yet somehow $\sigma^{12}(1) \neq 1$ , which is impossible. So $\sigma^2(6)$ must be $4$ instead.

This lets us decompose $\sigma$ into two odd-length cycles:

$\sigma = (1,12,7,6,11,4,8,10,3)(2,9,5)$

And applying the permutation to the input string $\text{GOLDENBINARY}$ yields the intermediate string $\text{LEARNBYDOING}$ , which can be represented in the desired format as $\boxed{120501181402250415091407}$ .