Unitary
\(U^\dagger = U^{-1}\).
They are often called isometries (symmetries of the system).
Theorem
An operator \(U\) is unitary iff it preserves the inner product of all vectors. I.e. if we define \(|V\rangle=U|v\rangle\) and \(|W\rangle=U|w\rangle\) then \((|v\rangle,|w\rangle)=(|V\rangle,|W\rangle)\).
In short, it maps orthonormal sets to orthonormal sets.
Proof. Assume \(U\) is unitary. Then \(\langle V|W\rangle = \langle Uv|Uw\rangle = \langle v|U^\dagger U|w\rangle = \langle v|\mathbb{I}|w\rangle = \langle v|w|rangle\).
Assume \((|V\rangle,|W\rangle) = (|v\rangle,|w\rangle)\). Then \(\langle Uv|Uw\rangle = \langle v|U^\dagger U|w\rangle = \langle v|w\rangle\). So, \(U^\dagger U = \mathbb{I}\). Then, \(U^\dagger U=\mathbb{I}\). Also, \(\langle j|i\rangle = \langle Uj|Ui\rangle = \langle j|U^\dagger U|i\rangle = \langle j|i\rangle = \delta^i_j = \delta^j_i = \langle i|j\rangle = \langle j|i\rangle^* = \langle j|(U^\dagger U)^\dagger|i\rangle^*\). Thus, \(U^\dagger = U^{-1}\). Hence, \(U\) is unitary.
Examples
Explicit example
\(U=\frac{1}{\sqrt{2}} \begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix}, UU^\dagger=U^\dagger U=\mathbb{I}\).