Thinking inside the box for once:
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,-----------,---------------,----------,
| 3 7 6 | 8 9 5 | 4 2 1 |
| 5 4 8 |*13 *12 *23 | 6 7 9 |
| 1 2 9 |#67 #47 #46 | 5 8 3 |
:-----------+---------------+----------:
|#6 89 25 |a35 a248 a234 | 39 *1 #7 |
|#7 89 12 |c1+36 128 b6+23|*9–3 5 #4 |
|#4 3 15 | 157 d7+1 9 | 8 #6 *2 |
:-----------+---------------+----------:
| 8 1 3 | 9 #6 #7 | 2 4 5 |
| 9 6 4 |*2 5 *1 | 7 3 8 |
| 2 5 7 |#4 *3 8 | 1 9 6 |
'-----------'---------------'----------'
467b5 form a so-called open triple, i.e. in each minirow and minicolumn, there is exactly one of them. So, they appear on a diagonal (like p267).
Moreover, the vertical permutation is 467 and the horizontal permutation has the same parity as 467 from b8, therefore, they appear on a falling diagonal.
Now consider 123. One of them appears in each minicolumn, so suppose the same is true for minirows.
Then the remaining two 123s in r5 are in c37, completing 132b6 vertically.
However, b8 horizontally has the permutation 231, which has the opposite parity, hence the 123 diagonal in b5 would be rising, so it would intersect the 467 falling diagonal, contradiction.
Therefore, 123b5 are not on a diagonal, and there is a row containing (at least) two of them.
This already cannot be r6, and if it were r4, we'd get 23b5p123, 6b5p6, 1b5p4, and 7b5p8, placing 67 on a rising diagonal, contradiction.
So, the repeat is in r5 completing a triple with r5c3. –3r5c7, stte
One thing which imo deserves a honourable mention is that since 467 appear on a diagonal and 123 don't, 589 cannot either.
This gives us (12=8)r25c5 –– 5r4c4 = (5–7)r6c4 = 7r6c5 => –1r6c5, leading to the same state as JCO's first step.