This thread is a record.
Given a Hilbert space $\mathcal{H}$.
Consider a normal operator: $$N:\mathcal{D}\to\mathcal{H}:\quad N^*N=NN^*$$ and its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$
Regard a unitary transformation: $$U:\mathcal{H}\to\mathcal{K}:\quad U^*=U^{-1}$$
How to check that the transformed becomes: $$M:=UNU^{-1}=\int_\mathbb{C}\lambda\mathrm{d}UEU^{-1}(\lambda)=:\int_\mathbb{C}\lambda\mathrm{d}F(\lambda)$$
On
First show that if $E$ is a spectral measure then so is $UE(\cdot)U^*$.
Now note that for a spectral measure $UE(\cdot)U^*$ we have $$ \left<f, \int \limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^* g\right>= \int\limits_{\sigma(N)} z\ F_{f, g}(dz) \ \quad (f \in \mathcal{H}, g \in U \mathcal{D}),$$ where $F_{f, g}(\cdot):=\left<f, UE(\cdot)U^*g\right>$ is a countably additive measure on a Borel sigma-algebra over $\sigma(N)$ with total variation less or equal $\|f\|\|g\|$.
For any $f \in \mathcal{H}$ and $g \in U\mathcal{D}$ we have
$$\begin{align}\left<f, UNU^* g \right>&=\left<U^*f, NU^*g\right> = \left<U^*f, \left( \ \int\limits_{\sigma(N)}z \ E(\mathrm{d}z)\right)U^*g\right> \\&= \int\limits_{\sigma(N)} z \ \left<U^*f, E(\mathrm{d}z)U^*g\right>=\int\limits_{\sigma(N)} z\left<f, UE(\mathrm{d}z)U^*g\right>\\&=\left<f, \int\limits_{\sigma(N)}z \ UE(\mathrm{d}z)U^*g \right>. \end{align}$$
Hence, $UNU^*= \int_{\sigma(N)}z \ UE(\mathrm{d}z)U^*$.
It remains a spectral measure: $$F(A):=UE(A)U^{-1}:\quad F(\varnothing)=0\quad F(\Omega)=1$$
There are no domain issues since: $$\|UE(A)U^*\psi\|=\|E(A)U^*\psi\|$$
But a formal calculation shows: $$U^*\psi\in\mathcal{D}(N):\quad\langle\varphi,UNU^*\psi\rangle=\int\lambda\mathrm{d}\langle U^*\varphi,E(\lambda)U^*\psi\rangle=\int\lambda\mathrm{d}\langle\varphi,UE(\lambda)U^*\psi\rangle\quad(\varphi\in\mathcal{H})$$
Concluding that the identity holds!