Let function $f : \Bbb R^{n \times n} \to \Bbb R$ be defined by
$$f(X) := a^TXb+c^TXEXd$$
where $a, b, c, d \in \Bbb R^n$ and $n \times n$ matrix $E$ is symmetric and positive semidefinite.
Under what conditions on $a$, $b$, $c$, $d$ and $E$ is there a unique matrix $X$ such that $f(X)=0$?
This cannot always be true, of course. For instance, if $a=b=c=d=0$ then any matrix $X$ will satisfy $f(X)=0$.
When $n=1$, $a^TXb+c^TXEXd=0$ has only the trivial solution if and only if exactly one of $ab$ and $cdE$ is zero.
When $n=2$, the solutions to $a^TXb+c^TXEXd=0$ are never unique. In fact, one can always pick two nonzero vectors $u$ and $v$ such that $a^Tu=v^Td=0$. Then $X=uv^T$ is a non-trivial solution.