For a given symmetric and positive definite matrix $A \in \mathbb{R}^{n\times n}$ having its columns being a basis in $\mathbb{R}^n$ we have:
$$A(A^T A)^{-1} A^T$$ $$ = A(A^2)^{-1} A $$
being a projection matrix. The term $(A^TA)^{-1}$ represents a normalizing factor since the columns of $A$ are not an orthonormal basis. What happens if the normalizing constant is defined by another symmetric and positive definite matrix $B$, so we have:
$$A(B^T B)^{-1} A^T$$ $$ = A(B^2)^{-1}A^T $$
Does this new form mimic a projection matrix? Does it have eigenvalues close to 0 or 1?
Let $P_n$ denote the set of positive definite $n\times n$ real matrices. The set of matrices of the form you describe, $$ X=\{AB^{-2}A\mid A,B\in P_n\}, $$ is equal to $P_n$. Indeed if $A,B\in P_n$ then $$ (AB^{-2}A)^T=A^T(B^T)^{-2}A^T=AB^{-2}A $$ and $$ v^T(AB^{-2}A)v=(B^{-1}Av)^T(B^{-1}Av)>0 $$ for any nonzero $v$ (since $A,B$ are nonsingular). This shows $X\subseteq P_n$.
Conversely any $C\in P_n$ is diagonalizable via an orthogonal matrix, and has positive eigenvalues. That is, $$ C=O^TDO $$ where $O^TO=I$ and $D=\mathrm{diag}(d_1,\ldots,d_n)$ with $d_i>0$. Let $D'=\mathrm{diag}(\sqrt{d_1},\ldots,\sqrt{d_n})$. Let $A=O^TD'O$ and $B=I$. Then $$ AB^{-2}A=A^2=O^T(D')^2O=O^TDO=C. $$ This shows $P_n\subseteq X$.
So the only thing you can say about the eigenvalues of such a matrix is that they are all positive and real.