TL;DR: The possible strings of magical operations are:
(first) 143 -> 484 -> 968 -> 1837 -> 9218 -> 17347 (sixth)
(first) 89 -> 187 -> 968 -> 1837 -> 9218 -> 17347 (sixth)
Rigorous proof follows...
Let's first make a few starting observations regarding the left and right numbers for a given number (result) and the number it is created from by a magical operation (operand):
1) If the left number of the result is 1 then the left and right numbers of the operand must either sum to 10 or more, or sum to 9 with a 1 carried from the addition of any middle numbers.
2) If the left number of the result is 1 more than the right number, then the left and right numbers of the operand must sum to the right number of the result, with a 1 carried from the addition of any middle numbers.
3) If the left and right numbers of the result are equal, then the left and right numbers of operand must sum to one of them, and there is no carried 1 from the addition of any middle numbers.
Any combination of left and right numbers for the result that don't meet the above criteria can't be formed by any operand.
Starting with the sixth number as 17347, then we can say that the fifth number must be...
a 4-digit number, since there is no 5-digit number that can produce the sixth number after a magical operation. The only two single-digit numbers that can produce 17 are 8 and 9, so the fifth number has to be 9ab8. The 9 has to be on the left so the numbers that sum to 8 on the right (from the prior magical operation on the fourth number) can be incremented by a carried 1. We can also determine that a and b have to sum to 3, and that the carried one that creates the 9 will also be added to a, making a one larger than b. The only solution is a=2 and b=1.
This gives us the fifth number: 9218
This then leads us to the fourth number...
which also has to be a 4-digit number abcd. We know that a and d have to add to 8, and b and c have to add to 11, and are therefore different numbers. Let's first consider the possible values for a and d. The combinations 2bc6, 3bc5, 5bc3, 6bc2, and 7bc1 are immediately out since there are no combination of numbers in a 3-digit or 4-digit third number that could make those. If it's 4bc4, then a 4-digit third number would have to be either 1gh3, 2gh2, or 3gh1, and there are no combinations of g or h that could produce two different values for b and c (since there can be no carried 1).
The only option left for the fourth number is 1bc7. This would be created by a third number xyz where x and z would have to add to create a 7 on the right and a 1 on the left, and we know from above this would mean it's 9y8. The value for b could be either 7 if y+y+1 is less than 10, or 8 if y+y+1 is greater than 10. Since b+c must be 11, the only combination that works is b=8, c=3, and y=6.
This gives us the fourth number: 1837
As well as the third number: 968
Now onto the second number(s)...
which have to be a 3-digit number a8c, where a+c equals 8 and the middle number is 8 to produce a carried 1. As above, the options 286, 385, 583, 682, and 781 are out since there are no 2-digit or 3-digit first numbers that can result in those. We're left with 484 or 187.
This gives us two possibilities for a second number: 187 and 484
And now onto the first number(s)...
For a second number 484 the first number must be a 3-digit number a4c, where a+c is 4 and a must be strictly less than (not equal to) c. Only a=1 and c=3 satisfy this.
For a second number 187, only a first number 89 works, since 98 has a left number greater than the right.
This gives us two possibilities for a first number: 89 and 143
