If I understand the problem correctly, you could (in theory) form your "permuted" integers as follows:

- Form a vector of length n whose i-th entry is i.
- Convert the entries to strings (of decimal digits).
- Permute the entries
- Concatenate the permuted entries
- Convert the resulting string to an integer in base ten.

If the resulting integer is to be a cube, it has to be congruent to 0, 1 or 8 (mod 9).

However, any concatenation of the integers 1 to n is congruent to the sum of the integers 1 to n, or n(n+1)/2 (mod 9).

This gives the requirement that n be congruent to 0, 1, 4, 7, or 8 (mod 9). (No number n(n+1)/2 is congruent to 8 (mod 9).)

This is of course only a "constant factor" thing. I find the argument persuasive that, when n is "large enough" (and the possibility of cubes is not ruled out by congruences etc) then there will very likely be cubes among the "concatenated permuted" integers 1 to n.

I find the argument similarly persuasive for higher powers.

I'd call the question an "open question" rather than a "Conjecture." It's a numerical curiosity, but AFAIK does not have any theoretical significance.