Every *-isometric isomorphism of $B(\mathcal{H})$ keep compact operators? Let $\mathcal{H}$ be a Hilbert space, $B(\mathcal{H})$ denotes the $\mathcal{C}^*$-algebra consisting of bounded linear transformation on $\mathcal{H}$ ($*$ is the adjoint). Now consider a isometric *-isomorphism of $B(\mathcal{H})$ onto it self namely $\phi$. If $K$ is a compact operator, then $\phi(K)$ is also compact. Is this claim true? 

Here are my thoughts: (I guess it is right)
1.Maybe we can consider finite-rank operators. If we can prove $\phi$ keeps rank-one operators, then it will keep finite-rank operators, and thus keep compact operators by the density and continuity. 
2.The ideal $\mathcal{K}$ formed by all compact operators is a minimal nonzero closed ideal of $B(\mathcal{H})$, and $\phi(\mathcal{K})$ is a closed ideal of $B(\mathcal{H})$. So if we can prove $\phi(\mathcal{K})\cap \mathcal{K}\not= \varnothing$, then $\phi(\mathcal{K})=\mathcal{K}$ and the claim is true. 

I think this question is not too difficult, but I just sticked. It also helpful if you can give me some hints or references. Thanks.
 A: This is a corollary of the fact that every $*$-automorphism of $\mathbb B(\mathcal H)$ is inner, and the hint given by MaoWao can also be used to prove this.  I'll expand on this hint a bit.
Indeed, if $\phi$ is an automorphism of $\mathbb B(\mathcal H)$, then minimal projections are mapped to minimal projections under $\phi$.  Thus to each unit vector $\xi\in\mathcal H$, $\phi(\xi\otimes\xi)$ is also a minimal projection, hence is of the form $(u\xi)\otimes(u\xi)$ for some unit vector $u\xi\in\mathbb B(\mathcal H)$.  Then the map $\xi\mapsto u\xi$ extends to a unitary $u\in\mathbb B(\mathcal H)$, and it follows that $\phi=\operatorname{ad}(u)$.
A: Note that a $*$-isomorphisms is automatically isometric.


*

*If $p$ is a rank-one projection, then it is minimal (and viceversa). Then $\phi(p)$ is also a minimal projection, so it is also a rank-one projection. A selfadjoint compact operator $x$ is a norm-limit of linear combinations of rank-one projections, and this will be preserved by $\phi$. So $\phi$ maps compact selfadjoints to compact selfadjoints. Now any compact $x$ is, by the polar decomposition, $x=vr$, with $r$ positive. Since $r=(x^*x)^{1/2}$, it is compact. So $\phi(x)=\phi(v)\phi(r)$ is compact. 

*Much easier than the above, when $H$ is separable: since $\phi$ is a $*$-isomorphism, it preserves ideals. And $K(H)$ is the only nontrivial (closed, bilateral) ideal. 
A: Thanks for MaoWao’s hint. The answer is affirmative. Now I finish the details.
(When I type this, Aweygan gives his answer)

Observation 1: $\phi$ keeps the spectrum, and thus keeps the positive operators. Indeed, since $\phi$ algebraic isomorphism, $(\lambda-T)A=id$ iff. $(\lambda-\phi(T))\phi(A)=id$. So $\lambda\in \rho(T)$ iff. $\lambda\in  \rho(T)$. $\phi$ keeps the spectrum implies $\phi$ keeps the positive operators (a operator $A$ is positive iff. $(Ax,x)\geq 0$ iff. $\sigma(A)\subseteq \mathbb{R}_{\geq 0}$). 
Observation 2: for orthogonal projective operators $P$ and $Q$. $(Px,x)\leq (Qx,x)$ iff. $Ran(P)\subseteq Ran(Q)$. (easy exercise)
Observation 3: $\phi^{-1}$ is also a $\mathcal{C}^*$-isomorphism. So above still holds if we replace $\phi$ by $\phi^{-1}$.

Suppose $Q$ is a rank-one orthogonal projection. Then $$\phi(Q)^2=\phi(Q^2)=\phi(Q), \hspace{0.3cm} \phi(Q)^*=\phi(Q^*)=\phi(Q).$$ Thus $\phi(Q)$ is an orthogonal projection. To see $\phi(Q)$ is rank-one, it’s sufficient to see it is minimal. If there is some orthogonal projection $P$ such that $(Px,x)\leq (\phi(Q)x,x)$. By above discussion and observation 3, $\phi^{-1}(P)$ is orthogonal projection. Note $\phi(Q)-P$ is positive, use observation 1 with $\phi^{-1}$ yields $Q-\phi^{-1}(P)$ is positive. Then use observation 2 with $\phi^{-1}$ yields $Ran(\phi^{-1}(P))\subseteq Q$, hence $\phi^{-1}(P)=Q$ and $P=\phi(Q)$. This proves $\phi(Q)$ is minimal. Consequently $\phi$ keeps rank-one orthogonal projections.
Every (bounded) rank-one operator is a scalar of rank-one orthogonal projections, so $\phi$ keeps rank-one operators. And it follows by my strategy 1.
