If the $A$ is swapped 5 times, once with each of the other letters, the result will be $BCDEFA$. Now the $B$ can be swapped 4 times in the same way to end up in the fifth position: $CDEFBA$. Continuing in this way gives a sequence of $5 + 4 + 3 + 2 + 1 = 15$ swaps that achieves the required result.

To see that no sequence of fewer than 15 swaps will work, note that in $ABCDEF$ there are 15 instances of pairs of letters that are in alphabetical order ($AB$, $AC$, $AD$, $AE$, $AF$, $BC$, $BD$, $BE$, $BF$, $CD$, $CE$, $CF$, $DE$, $DF$, $EF$), and in the required final string there are no such pairs. Each swap can decrease the number of pairs of letters that are in alphabetical order by just 1, so at least 15 swaps are required.