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2 Took back a false claim.

Edit: This is just half an answer: I think can only show that the answer to your question is yes, even for higher powers but let me stick to squaressets matrices with $X^2=N$ and fixed Jordan type are path connected.

$(a_1-1,1,a_2-1,1,$(\lfloor (a_1 +1)/2 \rfloor, \lfloor a_1/2\rfloor ,\lfloor (a_2 +1)/2 \rfloor ,\lfloor a_2/2\rfloor, \ldots, a_r-1,1)$. \lfloor a_r/2 \rfloor)$$From here you we can derive a necessary and sufficient condition on a nilpotent matrix to be a square. Now fix your preferred nilpotent matrix N. Let X be a matrix with X^2=N and Jordan type (a_1, \ldots, a_r)where r is minimal. Conjugating the whole setup, we may assume X is in Jordan form. Let Y be a matrix with Y^2=X^2 = N and let us construct a path from X to Y. Let (b_1, \ldots, b_s) be the Jordan type of Y. By minimality of r we have b_i=a_i whenever b_i>1, and if b_i=1 we have a_i=1 or a_i=2. Look the 2\times 2 Jordan blocks of X: Replacing the 1 in appropriately many of those blocks by a t and varying t from 1 to zero yields a path inside the set of "square roots" \mathcal S_N from X to a matrix Z in Jordan form having the same Jordan type as Y. It remains to X, and let us construct a path from Z X to Y inside \mathcal S_N. Y. Since Z X and Y have the same Jordan type, there exists an invertible matrix S with Y=SZS^{-1}. Y=SXS^{-1}. Because Y^2=Z^2=N Y^2=X^2=N the matrix S commutes with N. It is enough to construct a path from the identity matrix to S in the set \mathcal C_N of invertible matrices that commute with N. I claim \mathcal C_N is path connected . (for just any N). Indeed, the set [N] of all commutators of N is linear subspace of the vector space \mathrm M_n(\mathbb C). The determinant, as a function on [N] is a polynomial function which is not identically zero since the identity matrix belongs to [N]. Thus, the zero set of the determinant is Zariski closed, so \mathcal C_N is Zariski open in [N]. Any Zariski-open in a complex vector space is path connected. What remains is to connect different Jordan types. We certainly can connect (5,2) with (5,1,1) by the 1 in the 2\times 2--block with t and vary t from 1 to zero. The problem that remains is to connect, say, type (4,2) with type (3,3) as pointed out in the comments. 1 I think the answer to your question is yes, even for higher powers but let me stick to squares. Every nilpotent matrix is conjugate to a nilpotent matrix in Jordan form, which is unique up to permutation of Jordam blocks. So we have a bijection$$\mathrm{Nilp}_n(\mathbb C)/\mathrm{conjugation} \quad \cong \quad \mbox{integer partitions of }n$$associating with a conjugacy class of a nilpotent matrix$X$the sizes of its Jordan blocks$(a_1, \ldots, a_r)$which sum up to$n$. The max of the$a_i$'s is the nilpotency-degree of$X$. To the class of$X^2$is associated the partition$(a_1-1,1,a_2-1,1, \ldots, a_r-1,1)$. From here you can derive a necessary and sufficient condition on a nilpotent matrix to be a square. Now fix your preferred nilpotent matrix$N$. Let$X$be a matrix with$X^2=N$and Jordan type$(a_1, \ldots, a_r)$where$r$is minimal. Conjugating the whole setup, we may assume$X$is in Jordan form. Let$Y$be a matrix with$Y^2=X^2 = N$and let us construct a path from$X$to$Y$. Let$(b_1, \ldots, b_s)$be the Jordan type of$Y$. By minimality of$r$we have$b_i=a_i$whenever$b_i>1$, and if$b_i=1$we have$a_i=1$or$a_i=2$. Look the$2\times 2$Jordan blocks of$X$: Replacing the$1$in appropriately many of those blocks by a$t$and varying$t$from$1$to zero yields a path inside the set of "square roots"$\mathcal S_N$from$X$to a matrix$Z$in Jordan form having the same Jordan type as$Y$. It remains to construct a path from$Z$to$Y$inside$\mathcal S_N$. Since$Z$and$Y$have the same Jordan type, there exists an invertible matrix$S$with$Y=SZS^{-1}$. Because$Y^2=Z^2=N$the matrix$S$commutes with$N$. It is enough to construct a path from the identity matrix to$S$in the set$\mathcal C_N$of invertible matrices that commute with$N$. I claim$\mathcal C_N$is path connected. Indeed, the set$[N]$of all commutators of$N$is linear subspace of the vector space$\mathrm M_n(\mathbb C)$. The determinant, as a function on$[N]$is a polynomial function which is not identically zero since the identity matrix belongs to$[N]$. Thus, the zero set of the determinant is Zariski closed, so$\mathcal C_N$is Zariski open in$[N]\$. Any Zariski-open in a complex vector space is path connected.