Let $R_1$, $R_2 \in \mathbb R^{N \times M}$ with $M > N$ be matrices of rank $N-1$. When can we find a symmetric matrix $P\in \mathbb R^{M\times M}$ such that the matrix equation
$$ R_1 P = R_2, $$
is satisfied? When does a solution exist? I was thinking that an approach could be rewriting the equation by vecotrizing $P$, but since it is a symmetric matrix, I don't know how to rewrite $R_1$ and $R_2$ accordingly. I implemented such system of equations in Matlab and it could not find a solution when $P$ is symmetric (even if there are more variables than equations), but if found a solution for a nonsymmetric matrix $P$.
Let $USV^T$ be a singular value decomposition of $R_1$. The equation can then be rewritten as $SP_1=R$, where $P_1=V^TPV$ is symmetric and $R=U^TR_2V$. Now partition $S$ and $R$ as $$ \pmatrix{\Sigma&0_{(N-1)\times(M-N+1)}\\ 0_{1\times(N-1)}&0_{1\times(M-N+1)}} \quad\text{and}\quad \pmatrix{X&Y\\ z^T&w^T} $$ respectively. It becomes clear that the equation $SP_1=R$ is solvable for a symmetric $P_1$ if and only if $\Sigma^{-1}X$ is symmetric and $z^T,w^T$ are zero vectors. When this necessary and sufficient is satisfied, the general solution is given by $P=VP_1V^T$ where $$ P_1=\pmatrix{\Sigma^{-1}X&\Sigma^{-1}Y\\ Y^T\Sigma^{-1}&Z} $$ and $Z$ is any $(M-N+1)\times(M-N+1)$ symmetric matrix. (In particular, you may take $Z=0$).
In terms of $R_1$ and $R_2$, this means the equation is solvable if and only if $(I_N-R_1R_1^+)R_2=0$ and $R_1^+R_2R_1^+R_1$ is symmetric. When this necessary and sufficient is satisfied, the general solution is given by $$ P=R_1^+R_2+(I_M-R_1^+R_1)(R_1^+R_2)^T+(I_M-R_1^+R_1)A(I_M-R_1^+R_1) $$ for any $M\times M$ matrix $A$ such that $(I-R_1^+R_1)A(I-R_1^+R_1)$ is symmetric. (In particular, you may take $A=0$.