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 $U'$.