Avoid answering in comments. For $2\times 2$ case only.
First we have an easy to prove lemma:
$$
\det(I+X)=\det(A)^2
$$
In our case:
$$
X = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} \quad ; \quad
A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}
$$
Then we have:
$$
\det(I+X) = (a_{11}a_{22}-a_{12}a_{21})^2
$$
And:
$$
X = A^TX(I+X)^{-1}A = \\
\begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}
\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}
\begin{bmatrix} 1+x_{22} & -x_{12} \\ -x_{21} & 1+x_{11} \end{bmatrix} / \det(I+X)
\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}
$$
After some tedious algebra, it follows that we have four non-linear equations with four unknowns.
Then we invoke MAPLE:
eqns := {
((a11*x11+a21*x21)*(x22+1)-(a11*x12+a21*x22)*x21)*a11+(-(a11*x11+a21*x21)*x12+(a11*x12+a21*x22)*(x11+1))*a21 = x11*(a11*a22-a12*a21)^2,
((a11*x11+a21*x21)*(x22+1)-(a11*x12+a21*x22)*x21)*a12+(-(a11*x11+a21*x21)*x12+(a11*x12+a21*x22)*(x11+1))*a22 = x12*(a11*a22-a12*a21)^2,
((a12*x11+a22*x21)*(x22+1)-(a12*x12+a22*x22)*x21)*a11+(-(a12*x11+a22*x21)*x12+(a12*x12+a22*x22)*(x11+1))*a21 = x21*(a11*a22-a12*a21)^2,
((a12*x11+a22*x21)*(x22+1)-(a12*x12+a22*x22)*x21)*a12+(-(a12*x11+a22*x21)*x12+(a12*x12+a22*x22)*(x11+1))*a22 = x22*(a11*a22-a12*a21)^2};
solve(eqns,{x11,x12,x21,x22});
assign(s[2]);
x11+x22;
The latter two statements because we are not interested in the zero solution and seek for the trace.
The end-result is:
$$
\mathrm{Tr}(X) = \frac {-2-2a_{11}a_{21}a_{22}^3a_{12}+a_{11}^2a_{22}^4-2a_{11}a_{22}a_{12}^3a_{21}
+4a_{11}a_{22}-2a_{12}a_{21}a_{11}^3a_{22}-2a_{11}^2a_{22}^2+2a_{12}^2a_{21}^2+a_{22}^2+a_{11}^2
-a_{21}^2-a_{12}^2 -2a_{21}^3a_{11}a_{22}a_{12}+a_{21}^4a_{12}^2+a_{12}^2a_{21}^2a_{22}^2
+a_{21}^2a_{11}^2a_{22}^2+a_{11}^2a_{12}^2 a_{21}^2+a_{11}^2a_{22}^2a_{12}^2+a_{12}^4a_{21}^2
+2a_{22}^2a_{21}a_{12}-2a_{22}^3a_{11}-2a_{22}a_{11}^3+a_{11}^4 a_{22}^2+2a_{11}^2a_{21}a_{12}}
{1+a_{21}^2+a_{11}^2a_{22}^2-2a_{11}a_{22}a_{12}a_{21}+a_{12}^2a_{21}^2+a_{12}^2-2a_{11}a_{22}}
$$