Let $A$ be a *-algebra and $a,b$ are unitarily equivalent in $A$ ( i.e. there exists a unitary $u$ of $A$ s.t $b=uau^{*}$ ). I want to prove that $\sigma(a)=\sigma(b)$, if $a,b$ are unitarily equivalent.
Also, I know that if $A$ is unital and $u$ is a unitary element in $A$ (i.e. $uu^{*}=u^{*}u=1$), then
$Ad~u:A\rightarrow A~,~a\mapsto~uau^{*}$
is an *-automorphism of $A$.
If $x$ and $v$ are elements of an algebra with identity $1$ and $v$ is invertible, then $x$ is invertible if and only if $vxv^{-1}$ is invertible. If $x^{-1}$ exists, then $(vxv^{-1})^{-1}=vx^{-1}v^{-1}$. The other direction follows because $x=v^{-1}(vxv^{-1})v$. (It could also be noted that $y\mapsto vyv^{-1}$ is an algebra isomorphism.)
Apply this with $x=a-\lambda 1$ and $v=u$ to see what happens.