= Solution
If $u$ is invertible, applying the <algebra involution> to $u^{-1}u=uu^{-1}=1$ gives
$$
u^*(u^{-1})^*=(u^{-1})^*u^*=1,
$$
so $u^*$ is invertible and $(u^*)^{-1}=(u^{-1})^*$. The converse follows by applying $*$ again. Since
$$
(x-\lambda1)^*=x^*-\overline\lambda1,
$$
we have $\lambda\notin\sigma_A(x)$ if and only if $\overline\lambda\notin\sigma_A(x^*)$. Taking complements and using that <complex conjugation> is a bijection gives
$$
\boxed{\sigma_A(x^*)=\{\overline\lambda:\lambda\in\sigma_A(x)\}.}
$$
The conjugation bar in the printed PDF is essential; this is not an assertion that the two spectra always coincide.
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