We have the following lemma: Let $B(H)_s$ be the set of self-adjoint operators on $H$,suppose $\psi :B(H)_s\to B(H)_s$ is a bijective map which preserves commutativity(i.e. $MN=NM$ iff $\psi(MN)=\psi(NM)$),,then there exits an unitary operator or an antinuitary $U$ s.t. for any operator $S\in B(H)_s$,there exists a real-valued bounded Borel function $g_S$ on $\sigma(S)$ s.t $\psi(S)=Ug_S(S)U^*$.
Question:Suppose $\phi $ is a bijective map on $B(H)_{ip}$ which preserves commutativity,where $B(H)_{ip}$ is the set of invertible positive opertaors on $H$.Construct a bijective map $h$ on $B(H)_s$ such that $h(T)=log(\phi(e^T)),\forall T\in B(H)_s$ and preserve commutativity.How to conclude that $\phi(A)=Uf_A(A)U^*,\forall A\in B(H)_{ip}$,where $f_A$ is a real-valued bounded Borel function on $\sigma(A)$.
Let me denote your $h$ by $\psi$. Then, by the lemma, $$ \log(\phi(e^T)) = \psi(T) = Ug_T(T)U^* $$ for any $T\in B_s$. Now, let $A\in B_{ip}$ and set $T=\log(A)\in B_s$ as well as $f_A(t) = e^{g_T(\log t)}$. Then $$ \phi(A) = \phi(e^T) = Ue^{g_T(T)}U^* = Uf_A(A)U^*. $$