Inversion reverses the order of positive definite matrices

Homeomorphism. If $S\in\mathrm{PD}$, its eigenvalues are positive, so $S^{-1}$ is symmetric with eigenvalues $1/\lambda_i>0$: $S^{-1}\in\mathrm{PD}$. The map is continuous on $GL_n$ (entries of $S^{-1}$ are $\operatorname{adj}(S)/\det S$, rational with nonvanishing denominator), it is its own inverse, hence a homeomorphism of $\mathrm{PD}$.

Order reversal. Let $0<A\le B$ and let $C=A^{-1/2}$ (symmetric, positive definite). Congruence preserves the order: $X\le Y\Rightarrow CXC\le CYC$, because $v^TC(Y-X)Cv=(Cv)^T(Y-X)(Cv)\ge0$. Thus
$$I=CAC\le CBC=:T.$$
So all eigenvalues of the PD matrix $T$ are $\ge1$, hence the eigenvalues of $T^{-1}$ are $\le1$, i.e. $T^{-1}\le I$. Now $T^{-1}=C^{-1}B^{-1}C^{-1}=A^{1/2}B^{-1}A^{1/2}$, so
$$A^{1/2}B^{-1}A^{1/2}\le I.$$
Conjugating by $A^{-1/2}$ again: $B^{-1}\le A^{-1}$.

Example. $A=I$, $B=\operatorname{diag}(1,4)$: $B^{-1}=\operatorname{diag}(1,\frac14)\le I=A^{-1}$.

Remark. The map $S\mapsto S^{-1}$ reverses the order but $S\mapsto S^2$ does *not* preserve it (while $S\mapsto\sqrt S$ does, see the operator monotonicity problem). A function $f$ preserving the Loewner order is *operator monotone*; $-1/x$, $\ln x$, $x^{r}$ ($0\le r\le1$) are, and $x^2$ is not.