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 let $t$ be a word starting with $k$, not square-initial. For $m\ge1$ and $s<_{lex}t$, call $\ tst.tst.s^m$ an S-word, e.g. t=10, s=0, m=1 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 there square-initial Nyldon words not fitting this pattern?

<strike>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.  </strike>

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.