This seems a shorter solution:

def findMove(original,result):
    n = len(original)
    if n != len(result):
        return # different length                                               

    # find common head
    i=0
    while i<n and original[i] == result[i]:
        i += 1
    if i==n:
        return # equal strings                                                  

    # find common tail
    j=n-1
    while original[j] == result[j]:
        j -= 1

    # two possibilities: i->j or j->i
    if original[i+1:j+1] == result[i:j] and original[i] == result[j]:
        return (i,j)
    if original[i:j] == result[i+1:j+1] and original[j] == result[i]:
        return (j,i)

This seems a shorter solution:

def findMove(original,result):
    if len(original) != len(result):
        return # different length                                               

    diff = [x for x,(c,d) in enumerate(zip(original,result)) if c!=d]
    if not diff:
        return # equal strings                                                  
    i,j = diff[0],diff[-1]

    if original[i+1:j+1] == result[i:j] and original[i] == result[j]:
        return (i,j)
    if original[i:j] == result[i+1:j+1] and original[j] == result[i]:
        return (j,i)

PS: this works also if the items are non unique.