No, a bubble op bubbles one element into the correct position in a sorted list. If it was restricted to bubbling the maximum or minimum to the end of the list, the full sort would use one bubble, it would complete in linear time, and the list wouldn't be sorted afterwards.

You're correct that this is also a good match to insertion sort, but only conceptually.

(Note that one classic bubblesort pass moves several elements when the data is unsorted - when an element has bubbled as far as it can go, the bubblesort pass picks up the element that blocked it and carries that one further, and this process repeats until the end of the data is reached. For our purposes, there is no difference, because we're bubbling into a perfectly sorted list, so once our initial element stops moving, nothing else can move.)

The bigger difference between an insertion op and a bubble op is in the number of assignments.

This is the chain of operations you get while bubbling:

    swap(a5, a6)
    swap(a4, a5)
    swap(a3, a4)

This is the chain of operations you get while inserting:

    temp = a6
    a6 = a5
    a5 = a4
    a4 = a3
    a3 = temp

Inserting never does any swaps. But each swap is three assignments, so the bubble in this case does nine assignments to the insertion's five. (And in general, if you need to sink a value n places into the data, bubbling it there will take 3n assignments, and inserting it there will take n+2.) That is the advantage of knowing where you're going before you get there.