Consider $A, B \in \mathbb{R}^{n \times n}$, $A$ invertible, $B \succ 0$.
Say if the following holds: $$ A^{- \top} B A^{-1} \preccurlyeq B \ \Longleftrightarrow \ B \preccurlyeq A^\top B A. $$
I do not know if I can multiply the LHS by $A^\top$ on the left and by $A$ on the right. For sure this is fine if $A = a I$, $a \neq 0$.
Comment: $B \succ 0$ means that $B$ is positive definite, i.e. $B = B^\top$ and $B$ has positive eigenvalues.
I tried to say that: (i) the claim is true for $B = I$; (ii) $B = C^{-\top} C^{-1}$ for some $C \succ 0$.
Hint. Back to the basics. $P\succeq Q$ means $P-Q\succeq0$, i.e. $x^\ast(P-Q)x\ge0$ for every vector $x$.