Consider the set of $d\times d$ matrices that are positive semidefinite and have unit trace. This is a convex set, $S$. Is it possible to think of a geometric center of this set? The criterion for geometric center is an element $\rho$ such that if $\rho + X\in S$, then $\rho - X\in S$
For $d = 2$, it is the identity matrix. For $d>2$, it is not the identity matrix. Indeed $$\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix} + \begin{pmatrix} 2/3 & 0 & 0\\ 0 & -1/3 &0 \\ 0 & 0 & -1/3 \end{pmatrix} \in S$$ but $$\frac{1}{3}\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 2/3 & 0 & 0\\ 0 & -1/3 &0 \\ 0 & 0 & -1/3 \end{pmatrix} \notin S$$
So is it possible to find such an element for $d>2$ or if not, why not?
When $d\ge3$, there cannot be any geometric centre. Since the trace of a matrix and the property of being a centre are preserved by unitary similarity, we may consider only the case where $\rho$ is a nonnegative diagonal matrix.
If $\rho=\frac1nI$, then a similar example to yours shows that $\rho$ is not a centre.
If $\rho$ contains two different diagonal entries $a>b$ instead, $\rho$ is also not a centre because $$ \pmatrix{a\\ &b\\ &&c\\ &&&\ddots} +\pmatrix{-\frac{a+b}2\\ &\frac{a+b}2\\ &&0\\ &&&\ddots} =\pmatrix{\frac{a-b}2\\ &\frac{a+3b}2\\ &&c\\ &&&\ddots}\succeq0 $$ but $$ \pmatrix{a\\ &b\\ &&c\\ &&&\ddots} -\pmatrix{-\frac{a+b}2\\ &\frac{a+b}2\\ &&0\\ &&&\ddots} =\pmatrix{\frac{3a+b}2\\ &\frac{b-a}2\\ &&c\\ &&&\ddots}\nsucceq0. $$