I have a complex matrix $X$ whose size $m$ x $n$ where $m<n$ (for example, $m = 2$ and $n = 6$), I need to get the invertible real matrix $Y$ whose size $n$ x $n$ representing the null space of the imaginary part of matrix $X$. That means that $X\times Y$ is real matrix.
I think the first step of the solution is to get the null space matrix $Y_1$ for the imaginary part of $X$. However, by definition, the rank of the null-space matrix $Y_1$ is less than $n$, in my case, it's $5$. So, does it exist a real column $v$ such that $Y = [Y_1\ \ \ v ]$ is a real and invertible matrix? ($v$ can be a column or a matrix following the size of $Y$ which must be $n$ x $n$).
I don't know the answer, but maybe this might help you
Any complex matrix $X$ can be partitioned to its real and imaginary part as the following:
$$X=\frac{X+X^*}{2}+\frac{X-X^*}{2},$$
where $\frac{X+X^*}{2}$ is real and $\frac{X-X^*}{2}$ is complex.
So you need to find invertible real $Y$ such that
$$\Big(X-X^*\Big)Y=0.$$
Lets say that $X-X^*=[i\quad 2i]$ and $Y=\begin{bmatrix} y_{11} & y_{12}\\ y_{21}&y_{22} \end{bmatrix}$, then we will get:
$$y_{11}+2y_{21}=0 \qquad\Longrightarrow \qquad y_{11}=-2y_{21}$$ $$y_{12}+2y_{22}=0 \qquad\Longrightarrow\qquad y_{12}=-2y_{22}$$
then $\det(y_{11}y_{22}-y_{12}y_{21})=\det(-2y_{21}y_{22}+2y_{22}y_{21})=0$, thus $Y$ is not invertible. So I guess you cannot find $Y$, at least not for general $n$.
EDIT: Let $Z=\Im(X)$, then for your case, $\mathrm{dim}[\mathrm{null}(Z)]=5$, which means you cannot construct invertible $6\times 6$ matrix $Y$, such that $ZY=0$, but you can construct $6\times 5$ matrix $Y_1$, such that $ZY_1=0$ or singular $6\times 6$ matrix $Y$, such that $ZY=0$.