I really like this query! Here is what I managed to consider. The puzzle assertion makes use of minimal math as required. I’m much less positive whether or not it counts as a “good” puzzle in its present state, I believe it’s kind of too lengthy and convoluted. But hopefully somebody extra skilled in puzzle-making can take it as a place to begin, or at the very least discover some attention-grabbing concept they’ll use. (I additionally apologize prematurely for my English, I’m a non-native speaker).

Puzzle:

A bunch of evil aliens have at their disposition a robust spaceship fleet, which they use to destroy planets every time they really feel like.

The fleet consists of 100 ships in complete, numbered from $1$ to $100$. It is thought that the aliens use a set of connectors, full of a really highly effective however very unstable chemical substance, to energy up these ships. This chemical flows via plenty of cables, which be part of the inputs of the connector to its outputs in some particular approach. The cables cannot be seen from the surface, however every connector comes with a corresponding label depicting what’s inside.

The circuit board of Ship $N$ makes use of a set sequence of chained $N$-cable connectors, with the $N$th general enter of the sequence linked to the $N$th respective general output as within the image under.

The circuit is at all times designed in such a approach that the chemical flows in a single closed loop. This is extraordinarily essential, in any other case the ensuing magnetic instability would make the ship explode.

The aliens collectively enter a deep hibernation state each evening. There is a gaggle of saboteurs whose house planet is subsequent in line to be destroyed, and so wish to disable the alien fleet earlier than that may occur. They plan to do that by swapping the labels of the connectors in the correct approach whereas the aliens are sleeping.

But the aliens are conscious of the opportunity of a sabotage: in any case, they’ve garnered many enemies throughout their planet-destroying quest. As talked about, the connectors include a particularly unstable chemical, so that they can’t be opened or tampered with in any respect. They cannot even verify the inputs one after the other: the chemical should at all times move via all of the cables on the similar time in order to not disturb the fragile equilibrium of forces. Despite these limitations, the aliens have of their possession the Connector Chain Comparing Contraption (abbreviated C4), a miracle of engineering which is ready to evaluate any two arbitrary chains of connectors and inform whether or not they act the identical for all attainable enter alerts or not. This permits the aliens to make it possible for the labels of the connectors are internally constant. For instance, the C4 can be utilized to verify whether or not two copies of the three-cable connector labelled XI chained collectively act the identical as a single three-cable connector labelled III; if the C4 returns a “No” reply, it’s sure that at the very least one among these three connectors is mislabeled. To forestall the ships from being harmed, the aliens at all times use the C4 to meticulously check each attainable pair, triplet, and many others. of $N$-cable connectors (not solely those that will probably be loaded onto the ship, however the entire provide) for inconsistencies within the labels earlier than loading the connector chain onto Ship $N$‘s circuit board.

However, the aliens know that the C4 is just not highly effective sufficient to make sure that the labels match the true content material of the connectors. Long in the past, there was an incident the place the producer despatched all of the connectors of Ship $57$ horizontally flipped by mistake (that producer is now useless). Clearly, the C4 is unable to detect that form of mistake. Fortunately for the aliens, such “C4-invisible” mislabelings have at all times turned out to be inoffensive in apply: in each identified case, the mislabeled sequence of connectors nonetheless produces a circuit with a single loop when loaded onto the ship. The following picture exhibits one other instance, the place a former group of saboteurs managed to bypass C4 by following the technique depicted within the higher a part of the picture: assign a given $4$-cable label to the connector obtained by transferring the third enter and output to the leftmost place (holding the connections the identical), and do that for all attainable $4$-cable labels. This sabotage failed; as proven within the decrease a part of the picture, the mislabeled chain of connectors nonetheless produced a single loop of cable, permitting Ship $4$ to perform usually.

After checking numerous examples by hand, the aliens have concluded that every one “C4-invisible” mislabelings are probably innocent, and consequently that their routine checks with the C4 gadget will suffice to stop any attainable sabotage. But are they proper…?

Can the saboteurs succeed? If so, what technique should they observe?

Solution:

An $N$-cable connector could be regarded as a map from the set of $N$ inputs (taken so as) to the set of $N$ outputs (taken so as). This evidently corresponds to a permutation, i.e., a component of the group $S_N$. The existence of the C4 machine forces the saboteurs to relabel these parts in such a approach that composition of permutations is preserved, so a “C4-invisible” label swapping defines an automorphism of $S_N$.

If we load an arbitrary chain of connectors to Ship $N$, generically we may have $n$ loops, with respective lengths $l_1, l_2, ldots, l_n$, such that $sum_{i=1}^n l_i=N$. The unordered record $(l_1, l_2, ldots, l_n)$ defines the cycle kind of the permutation related to the chain of connectors. The assertion of the puzzle implies that Ship $N$ will work if and provided that the circuit kinds a single loop, during which case the cycle kind is essentially $(N)$. It is thought that interior automorphisms of $S_N$ protect the cycle kind of its parts, so since $S_N$ has no outer automorphisms for $Nneq 6$, it’s mathematically unimaginable for the label swapping to make Ship N malfunction if $Nneq 6$.

However, if $N=6$, it may be simply checked that the outer automorphism of $S_6$ sends permutations of cycle kind $(6)$ into permutations of cycle kind $(1,2,3)$, i.e., it turns the one loop into three loops of various lengths. So that is the answer of the issue: swapping the labels of all of the $6$-cable connectors in keeping with this automorphism (i.e., in keeping with any consultant within the equivalence class similar to the nontrivial component of $mathrm{Out}(S_6)$) will make Ship $6$ explode and save the saboteurs’ house planet.

I do not understand how believable it’s that somebody who is not already accustomed to group idea may bump into the answer, however at the very least the emphasis on loops of cables naturally leads one to consider cycle sorts, which is an important component of most proofs that $S_6$ is the one symmetric group with a nontrivial outer automorphism (see e.g. right here).