New attempt after my comment to the OP (but this one is too long for a comment):
EDITED after comments from user38477.
Assume wlog $\mathfrak A=\{1,...,k\}$ and letLet $t$$r,s,t$ be a word starting withnon empty words such that $k$$tst$ is Nyldon, not square-initial. For $m\ge1$ and$tstr$ is $s<_{lex}t$not Nyldon, and $r<_{lex}tst$. Then call $\ tst.tst.s^m$$\ tst.tst.r$ an S-word,. e.g. t=10, s=0, m=1r=0 yield $10010100100$. Also include among the S-words all words $\ tst.tst.s^m$ where $t$ itself is an S-word. (Here it really starts getting messy...) Are Are there square-initial Nyldon words not fitting this pattern?
Assume wlog $\mathfrak A=\{1,...,k\}$ and let $w$ be a word on $\mathfrak A$ that doesn't start with $k$, i.e. $w<_{lex}k$. Then $kwkkwkw$ (compare with @IlyaBogdanov's $1011010$!) is Nyldon, and I think I can show that all square-initial Nyldon words (i.e. those starting with a square) are of this form. Call those S-words. Note that $w$ can contain $k$'s at non-ínitial places, those do not give any trouble.
The new conjecture would be:
For a given necklace, its Nyldon representative (exists and) is the biggest shift in alphabetical order, excluding all the square-initial shifts that are not S-words.