Let $a, b\in A_+$ be positive elements of some C*algebra $A$.
Assume furthermore that $a$ is invertible.
Is it true that
$$
\exists! x\in A_+\quad:\quad xax=b\quad ?
$$
Already in the case $A=M_2(\mathbb C)$, I don't know how to solve this.



I do not know about general C^{*}algebras, but the statement is true for complex matrices. Uniqueness: Assume b = xax. Then a^{1/2}ba^{1/2} = a^{1/2}xaxa^{1/2} = (a^{1/2}xa^{1/2})^{2}, which implies that a^{1/2}xa^{1/2} = (a^{1/2}ba^{1/2})^{1/2}. Since a is invertible, x must be a^{−1/2}(a^{1/2}ba^{1/2})^{1/2}a^{−1/2}. Existence: It is easy to check that $x=\sqrt{a}^{1}\sqrt{\sqrt{a}\hspace{.15cm}b\sqrt{a}\quad}\sqrt{a}^{1}$ satisfies the condition. I do not think that anything in this argument depends on the fact that we are considering matrices, but let me avoid claiming things about the subject which I do not know well. 


In the case of $\Bbb{M}_n(\Bbb{C})$, you should diagonalize $a$, say $d(\lambda_1,\cdots,\lambda_n)$. Then $a^{1/2}$ is $d(\sqrt{\lambda_1},\cdots,\sqrt{\lambda_n})$. Easily you can find $a^{1/2}$ (since $a^{1/2}$ is of course invertible). The rest is just following the previous answer: $x=a^{1/2}(a^{1/2}ba^{1/2})^{1/2}a^{1/2}.$ Then you can return the basis to the previous one (i.e. the basis before diagonalization of $a$). 

