Let $\mu_1,\mu_2,\ldots,\mu_k \in \Delta^{d-1}$ be $k\ (k\geq 2)$ distinct vectors on the standard simplex, where 
$$\Delta^{d-1} = \{\mu\in R^{d}:\|\mu\|_1 = 1,\mu_j \geq 0\}.$$
(With a slight abuse of notation, here $\mu_j$ means the $j$-th entry of $\mu \in R^d$.) We define $\Delta^{k-1}$ similarly. 

Our goal is to find a $\mu^* \in \Delta^{d-1}$ that satisfies $\mu^* \notin \{\mu_1,\ldots,\mu_k\}$, such that for some $w \in \Delta^{k-1}$,  
$$
\sum_{i} w_i \mu_i = \mu^*,
$$
and 
$$
\sum_{i} w_i \mu_i^{\otimes 2} = (\mu^*)^{\otimes 2}.
$$
Equivalently, one can think of constructing $\mu_1,\mu_2,\ldots,\mu_k$ for a given $\mu^*$.

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 $\mu^* = (1/d,\ldots,1/d)^\top$, it seems that the solution doesn't exist. So what about general settings?