Let $A$ be a semi-simple algebra over $\mathbb{C}$. An idempotent $x$ in $A$ is an element of $A$ such that $x^2 = x$. A minimal non-zero idempotent $y$ of $A$ is a non-zero idempotent of $A$ such that for each non-zero idempotent $x \in A$ satisfying $xy = x$ we have $x = y$. We have a linear 'transposition' map on $A$ that sends an element $x \in A$ to $x^T \in A$ with the following properties:
$1^T = 1$;
$(x^T)^T = x$ for all $x \in A$;
$(x y)^T = y^Tx^T$ for all $x,y \in A$.
Is it true that there exists a set of minimal idempotents $x_1,\dots,x_n$ in $A$ such that
$x_i^T = x_i$ for each $i \in 1,\dots,n$;
$\sum_{i=1}^n x_i= 1$;
each idempotent $y$ of $A$ such that $y^T = y$ can be written as $\sum_{j=1}^k x_{i_j}$ for some $k \leq n$ with $i_j \in 1,...,n$ for each $j \in 1,...,k$.
If $A$ is a matrix algebra and $T$ is the normal transpose on $A$, then the diagonal matrices with exactly one $1$ and $0$ elsewhere do satisfy this condition.
The product algebra $A=\Bbb{C}\times\Bbb{C}$ with $(x,y)^T:=(y,x)$ is a counter example since $A$ has only two minimal idempotents, $(1,0)$ and $(0,1)$ which are interchanged by $T$.
I expect the answer to be "yes" if you further imposed that $A\simeq\prod_{i=1}^rM_{n_i}(\Bbb{C})$ with distinct $n_i$: the "transpose map" ought to be of the form $x\mapsto g\cdot{}^tx\cdot g^{-1}$ for some $g\in A^\times$ and the usual "term by term transposition" $(x_1,\dots,x_r)\mapsto (~{}^tx_1,\dots,{}^tx_r) $.