Let $\vec{\mu}_1, \vec{\mu}_2,\ldots, \vec{\mu}_k \in \Delta^{d-1}$ be $k\ (k\geq 2)$ distinct vectors on the standard simplex, where $$\Delta^{d-1} = \{\vec{\mu}\in R^{d}:\| \vec{\mu}\|_1 = 1,\mu_j \geq 0\}.$$ (Here $\mu_j$ means the $j$-th entry of $\vec{\mu} \in R^d$.) We define $\Delta^{k-1}$ similarly.

Our goal is to find a $\vec{\mu}^* \in \Delta^{d-1}$ that satisfies $\vec{\mu}^* \notin \{\vec{\mu}_1,\ldots, \vec{\mu}_k\}$, such that for some $\vec{w} \in \Delta^{k-1}$,
$$ \sum_{i} w_i \vec{\mu}_i = \vec{\mu}^*, $$ and $$ \sum_{i} w_i \vec{\mu}_i^{\otimes 2} = (\vec{\mu}^*)^{\otimes 2}. $$ (Here $\vec{\mu}_i^{\otimes 2}$ means $\vec{\mu}_i\vec{\mu}_i^\top$.) Equivalently, one can think of constructing $\vec{\mu}_1, \vec{\mu}_2,\ldots, \vec{\mu}_k$ for a given $\vec{\mu}^*$.

Here is the question: Is there a principled way to do this construction? Or does there exist a solution at all?

In the simplest example where $k=2$ and $\vec{\mu}^* = (1/d,\ldots,1/d)^\top$, it seems that the solution doesn't exist. So what about general settings?

A variant of this problem is that we want to match $$ \sum_{i} w_i \vec{\mu}_i = \sum_{i} w_i^* \vec{\mu}_i^*, $$ and $$ \sum_{i} w_i \vec{\mu}_i^{\otimes 2} = \sum_{i} w_i^* (\vec{\mu}_i^*)^{\otimes 2}, $$ where $\vec{w}^*\in\Delta^{k-1}$, $\vec{w}^*\neq \vec{w}$, $\vec{\mu}_1^*,\ldots, \vec{\mu}_k^* \in \Delta^{d-1}$, and $\vec{\mu}_1^*,\ldots, \vec{\mu}_k^* \notin \{\vec{\mu}_1,\ldots, \vec{\mu}_k\}$.