Solve $ (X+A)(X+A)'=\mathrm{diag}(I,0)+AA' $ Let $A$ be a given $m \times n$ matrix with $m  \geq n$ and Rank$(A)=n$. Does there always exist  an $m \times n$ matrix $X$ (with Rank$(X)=n$) that solves,
$$
(X+A)(X+A)'=\left[\begin{array}{cc}I_{n} &  0_{n \times (m-n)} \\  0_{(m-n) \times n} &  0_{(m-n) \times (m-n)}\\ \end{array}\right]+AA'?
$$
Simplifying, I get 
$$
XA'+AX'+ XX'=\left[\begin{array}{cc}I_{n} &  0_{n \times (m-n)} \\  0_{(m-n) \times n} &  0_{(m-n) \times (m-n)}\\ \end{array}\right]
$$
but not sure where to go from here. 
I was trying to investigate the case where $m=n$ but didn't make progress.
I suspect the answer is yes (but this is pure speculation from the case when all objects are scalars). This reminds me of Sylvester equations https://en.wikipedia.org/wiki/Sylvester_equation and pseudo inverses https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse
 A: Let $Y=X+A$; then $YY^T=diag(I_n,0_{m-n})+AA^T=B$.
Since $rank(YY^T)\leq n$, if we want that a solution exists, then necessarily $rank(B)\leq n$.
EDIT 1. Since $AA^T$ is symmetric $\geq 0$, $rank(B)\geq n$ and we assume that $rank(B)=n$. Since $rank(AA^T)=n$, the spd symmetric matrix $AA^T$ is in the form $diag(U_n,0_{m-n})$, where $U$ is a pd  symmetric matrix. Finally we may assume that $B=diag(b_1,\cdots,b_n,0_{m-n})$ where $b_i>0$. We can choose $Y=\begin{pmatrix}diag(\pm\sqrt{b_i})\\0_{m-n,n}\end{pmatrix}$.
EDIT 2. Answer to user. Necessarily, for any $y$, $[0_n,y^T]B[0_n,y]=0$; then $AA^T$ is in the form $\begin{pmatrix}U_n&V\\V^T&0_{m-n}\end{pmatrix}$. Thus if $A=[P,Q]^T$,then $QQ^T=0$, that implies that $Q=0$. Finally $V$ is also a zero matrix and we are done.
Conclusion. If $A$ is not in the form $[P,0]^T$, then $0$ solution; otherwise the solutions are $Y=[F,0]^T$ with $FF^T=I_n+PP^T=Wdiag(b_i)W^T$ where $W\in O(n)$. Thus there are at least $2^n$ solutions: $X=\begin{pmatrix}Wdiag(\pm\sqrt{b_i})W^T\\0_{m-n,n}\end{pmatrix}-A$.
