Suppose $A$ is an invertible matrix with complex entries and satisfies $A^T = -A$. By the Jordan decomposition of $A$ it is seen that there exist at least one (and hence several) square roots of $A$, and by Hermite interpolation of the square root function there exists a distinguished square root of $A$ that is a polynomial in $A$. Up to this point, these results are irrespective of the anti-symmetric condition. Once we impose anti-symmetry, does there exist a square root $\sqrt{A}$ such that $\sqrt{A}^{T} = i \sqrt{A}$?
P.S. The existence of such a square root would yield a proof of a technical detail in my master's thesis, but there is another (would-be) result that would suffice, so I ask it here as well.
Given a complex-symmetric (invertible) matrix $A$ a result known as the Autonne-Takagi factorization gives the existence of a unitary $U$ such that $U^TAU$ is a positive diagonal. If instead $A^T = -A$, does there exist a unitary $U$ such that $U^T A U$ is a positive diagonal times $J$, where $J$ is a block diagonal with blocks \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
I have tried to prove this by simply adapting the proof of Autonne-Takagi that I know, but I have come up short on the last few steps. Any information will be greatly appreciated.
The answer to your first question is negative, unless $A=0$. For any matrix $B$, if $B^T=iB$, then $B=(B^T)^T=(iB)^T=iB^T=i(iB)=-B$ and hence $B$ is necessarily zero.
For your second question, we have $A=U(s_1J\oplus\cdots\oplus s_kJ\oplus0)U^T$ for some unitary matrix $U$ and some positive real numbers $s_1,\ldots,s_k$, where the zero diagonal sub-block is possibly empty when the size of $A$ is even. This is a well-known result, first proved by the number theorist L.K. Hua in the article On the theory of automorphic functions of a matrix variable $I$-geometrical basis, Amer. J. Math., 66(1944).