If $A$ is a C*-algebra and $x\in\text{Inv}(A)$ self-adjoint and $\|x\|=\|x^{-1}\|=1$. Is $a\mapsto xax^{-1}$ isometric?

Question

Let $A$ be a unital C*-algebra and $x\in A$ invertible and self-adjoint such that $\|x\|=\|x^{-1}\|=1$. Define a map $$\varphi_{x}\colon A\to A,\qquad \varphi_{x}(a):=xax^{-1}.$$ Is $\varphi_{x}$ an isometry? That is, $\|\varphi_{x}(a)\|=\|a\|$ for all $a\in A$.

I have proven that $x$ must be unitary and it is clear that $\|\varphi_{x}(a)\|\leq\|x\|\|a\|\|x^{-1}\|=\|a\|$. But I don't see why $\|a\|\leq\|\varphi_{x}(a)\|$.

Any suggestions are greatly appreciated!

Answer 1

Once you know that $x$ is unitary, you can proceed by showing that $\|ua\|=\|a\|$ for all unitaries $u\in A$ and all $a\in A$. Indeed, $$\|ua\|^2=\|a^*u^*ua\|=\|a^*a\|=\|a\|^2.$$ Similarly, we see that $\|au\|=\|a\|$, and thus $\|uau^*\|=\|a\|$.