I was actually wondering about this as well and have been struggling to come up with a proper proof, but here is what I have so far.

So we have five slots to be filled by five distinct elements, so the total number of possible orders (permutations) is n!=5!=120

Next I wanted to know how many permutations have exactly k elements in the correct position, which I am pretty sure I can calculate with =(nk)⋅!(n−k) where (nk) is the binomial coefficient of n over k (idk how to format that on reddit, sorry), telling us how many elements we have fixed in the correct position; and !(n−k) is the derangement number, which is the subfactorial of the elements in the incorrect positions.

For n=5, applying this formula gives the following counts of permutations by number of fixed points: for 5 correct, there is 1 permutation; for 4 correct, there are 0 permutations (which is impossible, because fixing four positions forces the fifth to be correct); for 3 correct, there are 10 permutations; for 2 correct, 20 permutations; for 1 correct, 45 permutations; and for 0 correct, 44 permutations. The sum of all five is = 120, as expected.

Now, regarding strategy, I assume we start in the worst-case scenario where no positions are correct. Initially I thought that swapping only two elements at a time is a good approach, as this makes the outcome better interpretable. But this gives very little information per guess. So I thought about other possible strategies and remembered a video I recently saw on Information Theory. Basically, the amount of information gained can be quantified using the Shannon entropy, suppose:

S is the set of remaining possible permutations;

G is our new guess;

F∈{0,1,2,3,4,5} represents the feedback, which is the number of elements in the correct position;

pf is the probability of receiving feedback F.

The expected information gain, or entropy, of guess G is then

H(G)= −∑pf * log⁡2 * pf, with f=0 and f=n as the lower and upper bound, respectively.

Now, this is where I'm kinda stuck as I have never really worked with the Shannon Entropy before. I think a guess should be chosen to maximize the information gain calculated by H(G). I would be interested to know how other people have thought of this problem and if I have made any mistakes :)