Help Reorder the alphabet

Sure, that's an example I thought of as well, but how would you go about proving that it can't work in 24 steps?

For instance, you can't say each step can at most cover "fixing" one letter's position - consider $\text{dcba}$ , which can be done in 2 steps.

Correct. It is a familiar result in the theory of permutation groups that an $n$ -cycle can be written as the composition of $n-1$ transpositions, but no fewer.

The outline of the proof is as follows. First, each permutation of the alphabet can be viewed (uniquely) as the composition of $c$ disjoint cycles. The ordered alphabet corresponds to the identity permutation, which consists of 26 1-cycles: $(1)(2)(3)\cdots (26)$ .

Next, when composing a transposition $\tau = (a\ b)$ to a permutation $\sigma$ , there are two possibilities. If the elements $a,b$ belong to different cycles, the cycles are joined together into one, thus reducing the number of cycles from $c$ to $c - 1$ : $(a\ b)\circ(a\ x_1\ x_2\ \dots\ x_m)(b\ y_1\ y_2\ \dots\ y_n) = (a\ x_1\ \dots\ x_m\ b\ y_1\ \dots\ y_n).$ But if $a, b$ belong to the same cycle, it is now split into two, thus increasing the number of cycles from $c$ to $c+1$ : $(a\ b)\circ(a\ x_1\ \dots\ x_m\ b\ y_1\ \dots\ y_n) = (a\ x_1\ x_2\ \dots\ x_m)(b\ y_1\ y_2\ \dots\ y_n).$ Therefore each transposition changes the number of cycles by $\pm 1$ ; and to increase the number of cycles, transpose two elements that belong to the same cycle in the disjoint-cycle decomposition.

The shifted alphabet $\mathbf{bcd\cdots yza}$ corresponds to the 26-cycle $(1\ 2\ \dots\ 26)$ ; the ordered alphabet $\mathbf{abc\cdots xyz}$ corresponds to 26 1-cycles $(1)(2)\dots(26)$ ; to increase the number of cycles from 1 to 26, we need therefore at least 25 steps, and this minimum is accomplished if and only if every transposition exchanges two elements that belonged to the same cycle.