Let $G$ be a countable semigroup, and let $H$ be a finite-dimensional Hilbert space. Let $(T_g)_{g\in G}$ be a unitary representation of $G$ over $H$, i.e. for all $g\in G$, $T_g : G \to G$ is a unitary operator, and $T_g$ is also an isommetry. In addition, $(T_g)_{g\in G}$ is a $G$-action on $H$, meaning that $T_{g_1 g_2}=T_{g_1} \circ T_{g_2}$ for all $g_1, g_2 \in G$.
Is it true that the action is diagonalizable, i.e. is it true that there is an orthonormal set $\left\{f_1,\dots, f_k\right\}$ in $H$ such that for all $i\in \left\{1,\dots, \ell\right \}$, there is an eigenvalue $(\lambda^{(i)} _g)_{g\in G}$ such that for all $g\in G$ $T_g(f_i)=\lambda^{(i)} _g f_i$? If this is not the case, would it be true if we assumed that $G$ is a group and not just a semigroup?
Let $$U_1=\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix},\quad U_2=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$$ Then $U_1$ and $U_2$ are unitary and not diagonalizable simultaneously. Let $G$ be the group generated by $U_1$ and $U_2.$ Then $G$ is countable and not diagonalizable.