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 the 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. Consider the following partition of $X_0^{1/2}$
$$
X_0^{1/2} := \begin{bmatrix}a & b \\ b & c\end{bmatrix}
$$
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 then in this case we get that $X_0^{1/2}$ is singular, which is a contradiction since $X_0>0$.