Let $M$ denote the space of all complex $n\times n$ matrices. ($M$ is a complex vector space.)


Let $C\subseteq M$ be the set of all conjugate-normal matrices (i.e., matrices $A$ with $AA^\*=\overline{A^\*A}$).


Let $S\subseteq C$ denote all symmetric matrices. (Then $S$ is a subspace of $M$.)


Is there a subspace $X$ of $M$ such that $S\subseteq X\subseteq C$ and $X\neq S$?



Are symmetric matrices a maximal subspace of conjugate-normal matrices
Yes, the symmetric matrices are a maximal subspace in the conjugate-normal matrices.


Let $B$ be conjugate-normal and assume that $A+B\in C$ for all $A\in S$. Then 
\begin{align\*}
0 & = (A+B)(A+B)^\* - \overline{(A+B)^\*(A+B)}
\\ & = (A+B)(\overline A + B^\*) - (A + B^T)(\overline A + \overline B)
\\ & = AB^\* + B\overline A - A\overline B - B^T\overline A.
\end{align\*}
Let $B = [b\_{ij}]\_{i,j=1}^n$ and $A = E\_{11}\in S$, the matrix unit, then above becomes
$$
0 = \left[\begin{array}{cccc} \overline{b\_{11}} + b\_{11} & \overline{b\_{21}} & \cdots & \overline{b\_{n1}} \\ b\_{21} & 0 &&0
\\ \vdots&&&\vdots
\\ b\_{n1} & 0 &\cdots&0 \end{array}\right] - \left[\begin{array}{cccc} \overline{b\_{11}} + b\_{11} & \overline{b\_{12}} & \cdots & \overline{b\_{1n}} \\ b\_{12} & 0 &&0
\\ \vdots&&&\vdots
\\ b\_{1n} & 0 &\cdots&0\end{array}\right]
$$
which implies that $b\_{1j} = b\_{j1}$ for $2\leq j\leq n$. Repeat this process for $A = E\_{ii}, 2\leq i\leq n$ to get that $B$ is symmetric. 


