On the solvability of a matrix equation

Question

Let $\{C_i\}_{i=1}^N$ be a set of $n\times m$ real matrices of full column-rank and such that $\mathrm{Range}[C_1,C_2,\dots,C_N]=\mathbb{R}^n$, $\{P_i\}_{i=1}^N$ a set of $m\times m$ positive definite real matrices. Moreover, let $Q>0$ be a positive definite $n\times n$ real matrix such that $Q\in\mathrm{Range}\,\mathcal{C}$, where $\mathcal{C}$ is the linear operator $$ \mathcal{C}\colon A\mapsto \sum_{i=1}^N C_i A C_i^\top. $$

Consider the following equation in the unknown variable $X\in\mathbb{R}^{n\times n}$ $$\tag{$\star$}\label{a} \sum_{i=1}^N C_i(C_i^\top X C_i)^{-\frac{1}{2}}P_i(C_i^\top X C_i)^{-\frac{1}{2}}C_i^\top=Q. $$ (Here, $A^{\frac{1}{2}}$ denotes the principal matrix square root of $A\ge 0$.)

My question: Does \eqref{a} always admit a positive definite solution $\bar{X}>0$?

Notice that for the particular case $P_i=I_m$, $i=1,2,\dots,N$, \eqref{a} reduces to $$\tag{$\star\star$}\label{aa} \sum_{i=1}^N C_i(C_i^\top X C_i)^{-1}C_i^\top=Q. $$ In this case, I have some evidence to believe that the answer is in the affirmative. However, the general case is still obscure to me.

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EDIT. As Noah Stein correctly noticed in his answer below, the answer is negative for a general $Q>0$. However, I figured out that I forgot a fundamental assumption in my question, which now I added.

Answer 1

Let $N=n$, $m=1$, $P_i = 1$ for all $i$, and let the $C_i$ be the standard unit vectors. Then the left hand side of $(\star\star)$ is the matrix whose diagonal entries are the inverses of the diagonal entries of $X$ and whose off-diagonal entries are zero. So there is no solution, positive definite or otherwise, for $Q$ with zeros off the diagonal.