I was solving the Riccati equation $XAX=B$ where $A$ and $B$ are positive matrices, with $A$ being strictly positive.
Solving this, I did
$U'A^{1/2}XA^{1/2}UU'A^{1/2}XA^{1/2}U = U'A^{1/2}BA^{1/2}U$
$U'A^{1/2}XA^{1/2}U = (U'A^{1/2}BA^{1/2}U)^{1/2}$
$X = A^{-1/2}U(U'A^{1/2}BA^{1/2}U)^{1/2}U'A^{-1/2}$
where $U$ is a unitary matrix.
It's clear that $X = A^{-1/2}(A^{1/2}BA^{1/2})^{1/2}A^{-1/2}$ is a solution but why is it positive and furthermore the unique positive solution?
In general, if $A$ is positive and $Y$ arbitrary, then $Y^\ast A Y$ is positive: It is clearly symmetric (or Hermitian, if you're working over $\mathbb{C}$) and $\langle Y^\ast A Y\xi,\xi\rangle=\langle A(Y\xi),Y\xi\rangle\geq 0$. Also the square root of a positive matrix is positive by definition. This proves that the matrix $X$ from your question is positive.
Uniqueness actually follows from the computation in your question. Just note that for unitary $V$ and positive $Y$ you have $(V^\ast Y^{1/2} V)^2=V^\ast YV$, which implies $(V^\ast Y V)^{1/2}=V^\ast Y^{1/2}V$ (remember that $V^\ast Y^{1/2}V$ is positive). Thus the solution you got is $X=A^{-1/2}(A^{1/2}B A^{1/2})^{1/2}A^{-1/2}$.