Edit:
Given two real symmetric matrices ${\bf A}$ and ${\bf B}$, of size $n \times n$, such that their traces satisfies $trace({\bf A}) \geq trace({\bf B})$. I wish to find a general $n \times n$ matrix ${\bf C}$, such that $trace({\bf AC}) \geq trace({\bf BC})$.
Old question: Given a real symmetric matrix ${\bf A} \in {\mathbb{R}^{n \times n}}$, we have $A$ is positive semidefinite, i.e., ${\bf A} \succeq {\bf 0}$. I wish to find a matrix ${\bf C} \in \mathbb{R}^{n \times n}$, such that ${\bf AC}$ is also positive semidefinite, i.e., ${\bf AC} \succeq {\bf 0}$. Does such ${\bf C}$ exist? If yes, what conditions should it be met? [I guess we may need to consider the rank/positive-semidefiniteness of ${\bf C}$.]
You can deduce some conditions on $\mathbf C$ by picking some specific choices of $\mathbf A$ to see what happens.
For example, you can start by taking a matrix $\mathbf A$ with a single $1$ in some diagonal entry, and $0$ everywhere else. What is $\mathbf A\mathbf C$ in that case? When will this be positive semidefinite?
This narrows down the options for $\mathbf C$ considerably, but you will need to look at some more complicated matrices $\mathbf A$. For example, you could take nearly-zero matrices with the $2\times 2$ block $\begin{bmatrix}1 & -1 \\ -1 & 1\end{bmatrix}$: in some sense, these are the next-simplest positive semidefinite matrix.
Watch out, though: in general, $\mathbf A \mathbf C$ will not be symmetric, even if $\mathbf A$ is (and even if both $\mathbf A$ and $\mathbf C$ are). You can still technically say that a non-symmetric matrix is positive semidefinite, but this is not done very often; probably, what you want this to mean is that $\mathbf A \mathbf C + (\mathbf A \mathbf C)^{\mathsf T} \succeq \mathbf 0$.