Under what circumstances can $ALA^{-1}+BLB^{-1}=2 XLX^{-1}$ be solved for the linear operator $X$ only depending on $A$ and $B$ but not on $L$? For $A,B,L$ linear operators, when is there a linear operator $X\{A,B\}$ such that
$$ALA^{-1}+BLB^{-1}=2 XLX^{-1}$$
can be solved independently for all $L$ only depending on $A$ and $B$?
 A: For any fixed $k \neq 2$, there does not exist any $X \in GL(n,\mathbb{C})$ such that $$ A L A^{-1} + B L B^{-1} = k X L X^{-1}$$ for all $L \in M_n(\mathbb{C})$, for if there were, then taking $L=1_n$ yields
$$ 0 = A 1_n A^{-1} + B 1_n B^{-1} - k X 1_n X^{-1} = (2-k) 1_n,$$ which is impossible. In particular (unless I've misunderstood something), no such $X$ exists in your case, which is $k = \tfrac{1}{2} \neq 2$.

EDIT: In the case that $k = 2$, since every algebra automorphism of $M_n(\mathbb{C})$ is inner, and since $ALA^{-1} + BLB^{-1} = A(L + CLC^{-1})A^{-1}$ for $C = A^{-1}B$, your problem is equivalent to finding all $C \in GL(n,F)$ such that $L \mapsto \tfrac{1}{2}(L + CLC^{-1})$ is an algebra automorphism of $M_n(\mathbb{C})$. In the case that $C^2=1_n$, then you can check that this is true if and only if $C$ is a non-zero scalar multiple of the identity, in which case $\tfrac{1}{2}(L+CLC^{-1}) = L$, so that $B$ must be a non-zero scalar multiple of $A$, and you can just take $X=A$. Anything more general looks like a bit of a mess, though perhaps you can compute something explicit for $n=2$?
