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.

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 $\phi=\operatorname{ad}(u)$.
