I sketch here a solution for the $2\times 2$ case. My hope is that the outlined approach can be used to find a general solution.
As noticed in Addendum 1, we can restrict wlog to diagonal positive definite $A$'s $$ A := \begin{bmatrix}d_1 & 0 \\ 0 & d_2\end{bmatrix}. $$ I consider the case $d_1>1$ and $0<d_2<1$, the other cases being trivial. Define $X_0^{1/2}$ as $$ X_0^{1/2} := \begin{bmatrix}a & b \\ b & c\end{bmatrix}. $$ Since $X_0>0$ and hence $X_0^{1/2}>0$ it holds $a>0$, $b>0$, and $ac-b^2>0$. Moreover since $\mathrm{tr}(X_0)=1$, we have that $$ a^2+2b^2+c^2=1. \quad (1) $$ Now after one iteration step we obtain \begin{align} X_1 &= X_0^{1/2}AX_0^{1/2}= \begin{bmatrix} a^2d_1+b^2d_2 & \ast \\ \ast & b^2d_1+c^2d_2\end{bmatrix}. \end{align} Due to the fact that $\mathrm{tr}(X_1)=1$, it follows that $$ a^2d_1+b^2(d_1+d_2)+c^2d_2=1. \quad (2) $$ Let us define $$ X_1^{1/2} := \begin{bmatrix} a_1 & b_1 \\ b_1 & c_1\end{bmatrix} $$ Since $X_1=X_1^{1/2}X_1^{1/2}$, we get \begin{align} a_1^2+b_1^2=a^2d_1+b^2d_2 \ \ \text{ and }\ \ b_1^2+c_1^2=b^2d_1+c^2d_2. \quad(\#) \end{align} Now consider the second iteration step $$ X_2 = X_1^{1/2}AX_1^{1/2}= \begin{bmatrix} a_1^2d_1+b_1^2d_2 & \ast \\ \ast & b_1^2d_1+c_1^2d_2\end{bmatrix}. $$ We have \begin{align} \mathrm{tr}(X_2) &= d_1(a_1^2+b_1^2)+d_2(b_1^2+c_1^2)\\ &\overset{(\#)}{=}a^2d_1+2b^2d_1d_2+c^2d_2^2=1,\quad (3) \end{align} by virtue of $(\#)$ and of the trace constraint $\mathrm{tr}(X_2)=1$. By collecting $(1)$, $(2)$, and $(3)$ we arrive at the following linear system $$ \begin{bmatrix} 1 & 2 & 1 \\ d_1 & d_1+d_2 & d_2 \\ d_1^2 & d_1d_2 & d_2^2\end{bmatrix}\begin{bmatrix}a^2 \\ b^2\\ c^2\end{bmatrix}=\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}. $$ The solution $(\hat{a}^2, \hat{b}^2, \hat{c}^2)$ of the previous system — you can evaluate it manually or using some symbolic toolbox, as I did — is such that $$ \sqrt{\hat a^2 \hat c^2}-\hat b^2 = \frac{|(d_1-1)(d_2-1)|+(d_1-1)(d_2-1)}{(d_1-d_2)^2}=0 $$ since $d_1>1$ and $0<d_2<1$, by assumption. But, in this case, we get that $X_0^{1/2}$ is singular, which is a contradiction since $X_0>0$.
Update. It is possible to generalize the above procedure to the case $$ A = \begin{bmatrix}d_1I_{n_1} & 0\\ 0&d_2I_{n_2}\end{bmatrix} $$ with $d_1>1$ and $0<d_2<1$ scalars. Indeed, by considering the following block decomposition of $X_0^{1/2}$ $$ X_0^{1/2} := \begin{bmatrix}A & B \\ B^\top & C\end{bmatrix}, $$ by replacing $a$, $c$, and $b$ with $\mathrm{tr}(A^2)$, $\mathrm{tr}(C^2)$, and $\mathrm{tr}(B^\top B)$, and by following almost verbatim the above solution we get the desired result.