We know the standard form of SDP is
\begin{equation}\label{eq:ex_m} \begin{aligned} & {\underset{X}{\min}} & & \mbox{tr}(CX)\\ & \text{s.t.} & & \mbox{tr}(A_iX)=b_i, \ \ i=1,\cdots,p \\ & & & X\succeq0 \end{aligned} \end{equation}
Now, if I consider the following
\begin{equation} \begin{aligned} & {\underset{X}{\min}} & & \mbox{tr}(CX)\\ & \text{s.t.} & & \mbox{tr}(A_iX)=b_i, \ \ i=1,\cdots,p \\ & & & EXE\succeq0, \end{aligned} \end{equation}
where $E$ is a permutation matrix which permutes row and column. For example, permute row $i$ and row $j$, column $i$ and column $j$ with $i < j$. So $E$ is a orthogonal matrix. Note that, in this case, $E = E^{-1} = E^T$
I roughly ran a few examples, it seems that we can get the same cost. Not quite sure.
Q: Will we get the same cost and solution from both SDPs?
The optimal solutions (argmin) and optimal objective value are the same for both problems, presuming optimization is performed exactly.
That is because $EXE = E^{-1}XE$, which is similar to $X$. Therefore their eigenvalues are identical (see https://en.wikipedia.org/wiki/Matrix_similarity#Properties), so the constraints $X \succeq 0$ and $EXE \succeq 0$ are equivalent, in exact arithmetic.
If numerically solved in finite precision to finite optimality tolerance, it is possible for the solutions returned by a numerical optimizer to differ.