This is an expanded and corrected version of Subhasis's answer.
Formally, the problem is, given a n-letter alphabet V and two m-letter words, x and y, for which there exists a permutation p such that p(x) = y, determine the least number of swaps (permutations that fix all but two elements) whose composition q satisfies q(x) = y. Assuming that n-letter words are maps from the set {1, ..., m} to V and that p and q are permutations on {1, ..., m}, the action p(x) is defined as the composition p followed by x.
The least number of swaps whose composition is p can be expressed in terms of the cycle decomposition of p. When j1, ..., jk are pairwise distinct in {1, ..., m}, the cycle (j1 ... jk) is a permutation that maps ji to ji + 1 for i in {1, ..., k - 1}, maps jk to j1, and maps every other element to itself. The permutation p is the composition of every distinct cycle (j p(j) p(p(j)) ... j'), where j is arbitrary and p(j') = j. The order of composition does not matter, since each element appears in exactly one of the composed cycles. A k-element cycle (j1 ... jk) can be written as the product (j1 jk) (j1 jk - 1) ... (j1 j2) of k - 1 cycles. In general, every permutation can be written as a composition of m swaps minus the number of cycles comprising its cycle decomposition. A straightforward induction proof shows that this is optimal.
Now we get to the heart of Subhasis's answer. Instances of the asker's problem correspond one-to-one with Eulerian (for every vertex, in-degree equals out-degree) digraphs G with vertices V and m arcs labeled 1, ..., m. For j in {1, ..., n}, the arc labeled j goes from y(j) to x(j). The problem in terms of G is to determine how many parts a partition of the arcs of G into directed cycles can have. (Since G is Eulerian, such a partition always exists.) This is because the permutations q such that q(x) = y are in one-to-one correspondence with the partitions, as follows. For each cycle (j1 ... jk) of q, there is a part whose directed cycle is comprised of the arcs labeled j1, ..., jk.
The problem with Subhasis's NP-hardness reduction is that arc-disjoint cycle packing on Eulerian digraphs is a special case of arc-disjoint cycle packing on general digraphs, so an NP-hardness result for the latter has no direct implications for the complexity status of the former. In very recent work (see the citation below), however, it has been shown that, indeed, even the Eulerian special case is NP-hard. Thus, by the correspondence above, the asker's problem is as well.
As Subhasis hints, this problem can be solved in polynomial time when n, the size of the alphabet, is fixed (fixed-parameter tractable). Since there are O(n!) distinguishable cycles when the arcs are unlabeled, we can use dynamic programming on a state space of size O(mn), the number of distinguishable subgraphs. In practice, that might be sufficient for (let's say) a binary alphabet, but if I were to try to try to solve this problem exactly on instances with large alphabets, then I likely would try branch and bound, obtaining bounds by using linear programming with column generation to pack cycles fractionally.
@article{DBLP:journals/corr/GutinJSW14,
author = {Gregory Gutin and
Mark Jones and
Bin Sheng and
Magnus Wahlstr{"o}m},
title = {Parameterized Directed $k$-Chinese Postman Problem and $k$
Arc-Disjoint Cycles Problem on Euler Digraphs},
journal = {CoRR},
volume = {abs/1402.2137},
year = {2014},
ee = {http://arxiv.org/abs/1402.2137},
bibsource = {DBLP, http://dblp.uni-trier.de}
}