On Unitary Equivalent Observables

In J.J. Sakurai’s Modern Quantum Mechanics, he introduces the concept of ‘Unitary Equivalent Observables’.

If $|a^{‘}rangle$ and $|b^{‘}rangle$ are the orthonormal bases eigenkets of two non-commuting hermitian operators connected by the unitary operator $hat{U}$ ($|b^{‘}rangle = hat{U}|a^{‘}rangle$), we can construct a unitary transform of $hat{A}$ as $hat{U}hat{A}hat{U^{dagger}}$ such that $$hat{A}|a^{‘}rangle=a^{‘}|a^{‘}rangle$$ and $$hat{U}hat{A}hat{U^{dagger}}|b^{‘}rangle=a^{‘}|b^{‘}rangle.$$

However, the $|b^{‘}rangle$ satisfy: $$hat{B}|b^{‘}rangle=b^{‘}|b^{‘}rangle.$$

So, $hat{B}$ and $hat{U}hat{A}hat{U^{dagger}}$ are simultaneously diagonalizable.

My question is this. Are $hat{B}$ and $hat{U}hat{A}hat{U^{dagger}}$ the same operators? Sakurai’s own example using the spin operators $hat{S_{x}}$ and $hat{S_{z}}$ suggests so. Is there an example where they aren’t?

Also, what exactly does $hat{U}hat{A}hat{U^{dagger}}$ signify? As in, $hat{U}hat{A}hat{U^{dagger}}$ is the expression we’d get if we knew $hat{A}$ and $hat{U}$ in the $|b^{‘}rangle$ basis and wanted to know its matrix elements in the $|a^{‘}rangle$ basis.

$$langle a^{‘}|hat{A}|a^{”}rangle = langle a^{‘}|hat{U^{dagger}}hat{U}hat{A}hat{U^{dagger}}hat{U}|a^{”}rangle = langle b^{‘}|hat{U}hat{A}hat{U^{dagger}}|b^{”}rangle$$

How does that expression then give us an eigenvalue equation with $|b^{‘}rangle$? I think I’m missing something obvious here.