How to solve this quadratic matrix equation? I would like to solve for $X$ in the matrix equation
$$
XCX + AX = I
$$
where all the matrices are $n\times n$, have real components, $X$ is positive semidefinite and $C$ is symmetric.  My (possibly optimistic) hunch is that $X$ will be unique because of the positive semidefinite requirement; if not I only care about finding a single solution.
My searches have turned up a lot of work on solving non-symmetric riccati equations, unfortunately I don't meet the requirements for any of them.  For example, everything I've seen requires that the entries of $C$ all be nonnegative, which is not true in my case.
 A: In fact, your problem reduces to a Riccati equation in dimenion $n-1$ that is easy to solve. Here, if I correctly understood your problem, $A=I_n+aa^T, C=(cc^T)(cc^T)=uu^T$ (since $rank(C)=1$), where $a,u$ are known vectors. $A,X$ are symmetric $>0$ and $C$ is symmetric $\geq 0$, then $AX=XA$. We may assume $A=diag(1+\alpha,I_{n-1})$ where $\alpha\geq 0$. Then $X=diag(\beta,S_{n-1})$ where $\beta>0$ and $S$ is symmetric $>0$. Put $u=[v,w_{n-1}]^T$. $\alpha,v,w$ are known and $\beta,S$ are unknown. The equation can be rewritten: $\begin{pmatrix}\beta^2v^2&\beta vw^TS\\v\beta Sw&Sww^TS\end{pmatrix}+diag((1+\alpha)\beta,S)=I_n$. That implies $\beta^2v^2+(1+\alpha)\beta=1,vSw=0,Sww^TS+S-I_{n-1}=0$.
Case 1. (easy) $v\not=0,Sw=0$. Then $\beta^2v^2+(1+\alpha)\beta-1=0,S=I_{n-1}$. thus necessarily $w=0$ and for every $v$, there is a unique solution $\beta>0$.
Case 2. (more interesting) $v=0$. Then $(1+\alpha)\beta=1,Sww^TS+S-I_{n-1}=0$. You obtain $\beta>0$ if $\alpha>-1$ and finally you must to solve a standard symmetric Riccati equation in dimension $n-1$. As above, we may assume $ww^T=diag(\gamma,0_{n-1})$ where $\gamma\geq 0$. Then put $S=\begin{pmatrix}e&f^T\\f&H_{n-1}\end{pmatrix}$, where $e>0$ and $H$ is symmetric $>0$. The equation can be rewritten $(\gamma e+1)e=1,(\gamma e+1)f=0,H+ff^T=I_{n-1}$. Thus $f=0$ and $H=I_{n-1}$. For every value of $\gamma\geq 0$, there is a unique solution $e>0$.
EDIT: in fact, I assumed $\alpha>0$. It remains to study the case $\alpha=0$.
A: Both existence and uniqueness can fail spectacularly.
In the special case $A=0$ we get $XCX=I$, which is equivalent to $C = X^{-2}$.
Thus we must have $\det C > 0$, else then $C=X^{-2}$ has no real solution 
at all, even without requiring $X$ to be symmetric, let alone positive-definite.
If $X$ is symmetric then $C$ must also be positive-definite.
Note that $\det C$ can be zero or negative even when all entries of $C$
are positive.
When a solution does exist then there might even be a positive-dimensional
family of positive-definite $X$ satisfying the equation $XCX+AX=I$.
For example, let $A=3I$ and let $C=-2I$.
Then the equation becomes $2X^2-3X+I=0$, i.e. $(X-I)(2X-I)=0$.
So for any orthogonal decomposition ${\bf R}^n = V_1 \oplus V_{1/2}$
we get a solution by making $V_1$ the $1$-eigenspace
$V_{1/2}$ the $(1/2)$-eigenspace, i.e. letting $X$ be the transformation
that takes any $v_1+v_2$ ($v_i \in V_i$) to $v_1 + \frac12 v_2$.
A: Not a complete answer, but a long comment. Using standard results for Riccati equations, one can parametrize all (symmetric and non-symmetric) solutions. One can rewrite the equation in the form
$$
\mathcal{H}
\begin{bmatrix}
I\\ X
\end{bmatrix}
=
\begin{bmatrix}
I\\X
\end{bmatrix}CX, \quad \mathcal{H}=\begin{bmatrix}
0 & C\\
I & -A
\end{bmatrix}
$$
hence each solution is associated to an invariant subspace of the matrix $\mathcal{H}$. We can reverse that relation: (1) compute eigenvalues and eigenvectors of $\mathcal{H}$ (2) if the eigenvalues are all distinct, simply pick any $n$ out of $2n$ of them, and call them $u_1, u_2,\dots, u_n$ (with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$); otherwise the same holds with Jordan chain vectors, but you have to be careful not to take a vector in a chain unless you also take all the preceding ones (3) stack them horizontally to build $\begin{bmatrix}U_1 \\ U_2\end{bmatrix}=\begin{bmatrix}u_1 & u_2 & \dots & u_n\end{bmatrix}$, $U_1,U_2\in\mathbb{C}^{n\times n}$. (4) If $U_1$ is invertible, $X=U_2U_1^{-1}$ is a solution (which may or may not be symmetric), and $CX$ has eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$.
This gives all the solutions, which are at most $\binom{2n}{n}$ (in the nonderogatory case: if the matrix is derogatory, they are $\infty$ since there is arbitrariety in choosing the eigenvectors).
Typically the choice of these eigenvalues is "driven" by knowing which eigenvalues you want $CX$ to have. In your case you can at least say something: if $X$ has to be PD then $CX$ has all real eigenvalues and the same signature as $C$, and this restricts how you can choose the $\lambda_i$.
However, I am not sure that your matrix $\mathcal{H}$ has enough structure to say something general on its eigenvalues (let alone proving that there must be a symmetric solution -- as far as I know, this holds true only for some special classes of matrices, such as Hamiltonian or symplectic ones).
Can you run some experiments and tell us how the eigenvalues of $\mathcal{H}$ look like in your real-life cases? This could tell us something: for instance, if the matrix were symplectic, you would see them in pairs $(\lambda,1/\lambda)$.
