My solution don't use bilinear form.
Suppose that $X=\vec{A}+\vec{B}i= \left( \begin{array}{c} a_1+b_1 i \\ a_2+b_2 i \\ \vdots \\ a_n+b_n i \\ \end{array} \right), Y=\vec{C}+\vec{D}i= \left( \begin{array}{c} c_1+d_1 i \\ c_2+d_2 i \\ \vdots \\ c_n+d_n i \\ \end{array} \right).$ We have the following observation. If such matrix $B$ exists, suppose that $B=S+Ti= \left( \begin{array}{cccc} s_{11}+t_{11}i & s_{12}+t_{12}i & \cdots & s_{1n}+t_{1n}i \\ s_{21}+t_{21}i & s_{22}+t_{22}i & \cdots & s_{2n}+t_{2n}i \\ \vdots & \vdots & \ddots & \vdots \\ s_{n1}+t_{n1}i & s_{n2}+t_{n2}i & \cdots & s_{nn}+t_{nn}i \\ \end{array} \right).$ Since $BX=Y$, we have $\sum_{j=1}^{n}(a_j+b_j i)(s_{kj}+t_{kj}i)=(c_k+d_k i).$ Compare the real and imaginary parts, we get $\sum_{j=1}^{n}(a_j s_{kj}-b_j t_{kj})=c_k,$ $\sum_{j=1}^{n}(b_j s_{kj}+a_j t_{kj})=d_k.$ Then $\left( \begin{array}{cccc|cccc} \vec{A}^t & 0 & 0 & 0 & -\vec{B}^t & 0 & 0 & 0 \\ 0 & \vec{A}^t & 0 & 0 & 0 & -\vec{B}^t & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & \vec{A}^t & 0 & 0 & 0 & -\vec{B}^t \\ \hline \vec{B}^t & 0 & 0 & 0 & \vec{A}^t & 0 & 0 & 0 \\ 0 & \vec{B}^t & 0 & 0 & 0 & \vec{A}^t & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & \vec{B}^t & 0 & 0 & 0 & \vec{A}^t \\ \end{array} \right) \left( \begin{array}{c} S^t e_1 \\ S^t e_2 \\ \vdots \\ S^t e_n \\ T^t e_1 \\ T^t e_2 \\ \vdots \\ T^t e_n \\ \end{array} \right)= \left( \begin{array}{c} \vec{C} \\ \vec{D} \\ \end{array} \right), $ where $e_i$ is the $n\times 1$ column vector whose $i$ position is 1 and other 0. Write this equation by $MU=K$.
For example, when $n=2$, $M=\left( \begin{array}{cccc|cccc} a_1 & a_2 & 0 & 0 & -b_1 & -b_2 & 0 & 0 \\ 0 & 0 & a_1 & a_2 & 0 & 0 & -b_1 & -b_2 \\ \hline b_1 & b_2 & 0 & 0 & a_1 & a_2 & 0 & 0 \\ 0 & 0 & b_1 & b_2 & 0 & 0 & a_1 & a_2 \\ \end{array} \right).$ When $n=3$, $M=\left( \begin{array}{ccccccccc|ccccccccc} a_1 & a_2 & a_3 & 0 & 0 & 0 & 0 & 0 & 0 & -b_1 & -b_2 & -b_3 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & a_1 & a_2 & a_3 & 0 & 0 & 0 & 0 & 0 & 0 & -b_1 & -b_2 & -b_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & a_1 & a_2 & a_3 & 0 & 0 & 0 & 0 & 0 & 0 & -b_1 & -b_2 & -b_3 \\ \hline b_1 & b_2 & b_3 & 0 & 0 & 0 & 0 & 0 & 0 & a_1 & a_2 & a_3& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & b_1 & b_2 & b_3 & 0 & 0 & 0 & 0 & 0 & 0 & a_1 & a_2 & a_3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & b_1 & b_2 & b_3 & 0 & 0 & 0 & 0 & 0 & 0 & a_1 & a_2 & a_3 \\ \end{array} \right).$
We claim that the set of row vectors of $M$ are linearly independent. It is sufficient to show that the $l$-th row vector $v_l$ of $M$ and $(n+l)$-th row vector $v_{n+l}$ of $M$ are linearly independent for $l=1,2,...,n$. If $\lambda v_l+\gamma v_{n+l}=\vec{0}$, where $\lambda, \gamma\in \Bbb{R}$, then $ \begin{array}{c} \lambda a_1+\gamma b_1=0 \\ \lambda a_2+\gamma b_2=0 \\ \vdots \\ \lambda a_n+\gamma b_n=0 \\ \end{array} \mbox{ and } \begin{array}{c} -\lambda b_1+\gamma a_1=0 \\ -\lambda b_2+\gamma a_2=0 \\ \vdots \\ -\lambda b_n+\gamma a_n=0 \\ \end{array}.$ Since $X\neq 0$, there exists $i_0\in \{1,2,...,n\}$ such that $a_{i_0}\neq 0$ or $b_{i_0}\neq 0$. Then $a_{i_0}^2+b_{i_0}^2\neq 0$. Solve the equations $\lambda a_{i_0}+\gamma b_{i_0}=0,$ $-\lambda b_{i_0}+\gamma a_{i_0}=0.$ We get $\lambda(a_{i_0}^2+b_{i_0}^2)=0$ and $\lambda=0$. Similarly, we have $\gamma=0$. Hence $\lambda=\gamma=0$ and $v_{l}$ and $v_{n+l}$ are linearly independent.
Conversely, given $X\neq 0$ and $Y$, if we view the equation $MU=K$ as a linear system with $n^2$ unknowns in $U$, then since $rank(M)=2n=rank(M|K)$, by Theorem (p.174, theorem 3.11, Linear Algebra, 4 edition, Friedberg, Insel, Spence), $U$ has a solution.
Suppose that $U=\left( \begin{array}{c} s_{11} \\ \vdots \\ s_{ij} \\ \vdots \\ s_{nn} \\ t_{11} \\ \vdots \\ t_{ij} \\ \vdots \\ t_{nn} \\ \end{array} \right).$ For the symmetry of $B$, we need to require $s_{ij}=s_{ji}$ and $t_{ij}=s_{ji}$ in $U$. Hence we modify the vector $U$ into $U'$ such that there are $\frac{n(n+1)}{2}$ unknowns in $U'$. This modification reduces the column of $M$ into $M'$, but does not effect $rank(M)$ and $rank(M|K)$. That is, $rank(M')=rank(M)=rank(M|K)=rank(M'|K)$. Therefore, there is a solution for $U'$. Then we can construct the symmetry matrix $B$ from $U'$.