For vectors $0 \neq s,y \in \mathbb{R^n}$ with $(s,y)>0$ there exist only one positive definite symmetric matrix $Q$ such that $Qs = y$. The statement above was given as an exercise to prove. I've done the existence part. The problem with it is that the part about uniqueness might be a typo. Is it unique?
 A: This is clearly not true.
Take $s=y=e_1$ where $e_1$ is the first vector of the canonical basis then any diagonal matrix $Q$ with for diagonal $1, \lambda_2, \dots , \lambda_n$ where $\lambda_i \gt 0$ is symmetric positive definite and such that $Qs=y$.
A: No, it's not unique. Consider
$$s = y = \begin{bmatrix} 1 \\ 1\end{bmatrix}$$
Then there are infinitely many positive semi-definite matrices $Q$ that have $s$ as a $1$-eigenvector, namely $Q s = s = y$. To construct these examples, just consider an eigendecomposition of a positive definite $Q$ which has $s$ as a $1$-eigenvector. Then the eigendecomposition is of the form
$S D S^\dagger$
where $s / \|s\|$ is a column of $S$ (which is unitary), the corresponding diagonal entry in $D$ is one, and the rest of the diagonal entries in $D$ are also positive. For example,
$$\left( \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \right) \begin{bmatrix} 1 & 0 \\ 0 & \mu \end{bmatrix} \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\right)$$
satisfies the criterion for any positive $\mu$.
