Consider an irreducible non-Abelian subgroup $\mathrm{H}$ of group of unitary matrices $\mathrm{U}_n(\mathbb{C})$, that contains the subgroup of diagonal matrices. Does there exist any result regarding the properties or structures of $\mathrm{H}$? The final motivations is to answer this question.
P.S. The question has been updated after some useful comments.
1) First, let $H$ be a closed connected subgroup with this property. Let $D$ the diagonal group in $U(n)$; denote Lie algebras with Gothic letters. Then $$\mathfrak{u}(n)=\mathfrak{d}\oplus \bigoplus_{j<k}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj}).$$
Let $e_j:D\to\mathbf{C}^*$, $d\mapsto d_j$ be the projection (valued in the unit circle). As $D$-module, the above decomposition of $\mathfrak{u}(n)$ is invariant, $\mathfrak{d}$ has weight $e_j$ and $\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$ is $\mathbf{R}$-irreducible with 2-dimensional complexification of weights $\pm e_j-e_k$. Since these are distinct when the pair $(j,k)$ with $j<k$ varies, it follows that any $D$-submodule $M$ of $\mathfrak{u}(n)$ containing $\mathfrak{d}$ has the form $$M=\mathfrak{d}\oplus \bigoplus_{j<k;(j,k)\in W}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ for some subset $W$ of the set of pairs $(j,k)$ with $j<k$. Let $W'$ be the set of pairs $(j,k)$ such that $(j,k)$ or $(k,j)$ belongs to $W$. So $W'$ is symmetric and $$M=\mathfrak{d}\oplus \sum_{(j,k)\in W'}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ The condition that $M$ is a Lie subalgebra easily implies that $(j,k),(k,\ell)\in W'$ imply $(j,\ell)\in W'$. Hence $W''$, the union of $W'$ and the diagonal, is an equivalence relation on $\{1,\dot,n\}$. Conversely, for every equivalence relation $W''$ on $\{1,\dots,n\}$, $$\mathfrak{h}_{W''}=\mathfrak{d}\oplus \sum_{j\neq k;(j,k)\in W''}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$ is a Lie subalgebra containing $\mathfrak{d}$. The corresponding group is thus the group of block-diagonal matrices with respect to some partition (possibly permuting indices to make it block-wise).
2) Now let $H$ be a closed subgroup containing $D$, possibly not connected. Then $H^0$ has the preceding form, and $H$ normalizes $H^0$. One can check that the normalizer of $H^0$ is always of finite index over $H^0$: indeed, the blocks are precisely the irreducible components of the $H^0$-actions, and are pairwise non-isomorphic $H^0$-modules, hence they are permuted by $H$. That is, this normalizer is the stabilizer of some direct sum according to some partition of indices, possibly permuting blocks.
3) If one wants irreducibility (as I said in a comment, it just complicates the discussion while restricting the scope): it corresponds to the case where $H/H^0$ acts transitively on the set of blocks (this is possible only if all blocks have the same size)
4) The remaining step is to show that any subgroup $H$ containing $D$ is automatically closed. To start with, the connected component of the closure of $H$ is the component-wise stabilizer of some partition $P$ of $\{1,\dots,n\}$.
Let $x=(x_1,\dots,x_m)$ be an $m$-tuple of $H$. Consider the map $D^m\to U(n)$ mapping $(d_1,\dots,d_m)$ to $\prod x_id_ix_i^{-1}$. Let $r_x$ be its rank (maximum rank of its differential over $D^m$). So for some $y=(y_1,\dots,y_m)$, its rank at $y$ is $r_x$. Hence for $x'=(x_1,\dots,x_m,x_m,\dots,x_1)$, its rank at $(y_1,\dots,y_m,y_m^{-1},\dots,y_1^{-1})$ is $\ge r_x$ and moreover the value is $1$. Thus, we can assume that $x$ is chosen such that $r_x$ is maximal and achieved at a point $(y_1,\dots y_m)$ with value $1$. From the maximality it follows that the tangent image is a Lie subalgebra $\mathfrak{l}$ of $\mathfrak{u}(n)$, and it does not depend on the choice of $x$, and the corresponding immersed Lie subgroup $L$ is contained in $H$ and contains $D$. The preceding results concerns Lie subalgebras containing $\mathfrak{d}$ applies, so $\mathfrak{l}$ is the stabilizer of some partition $Q$ of $\{1,\dots,n\}$ (with $Q\subset P$ since $L\subset \bar{H}^0$). But it it easy to see that if $P\neq Q$ then $\mathfrak{h}_Q$ is not normalized by $\mathfrak{h}_P$. So $P=Q$. Hence $H\supset L=\bar{H}^0$. It follows that $H$ is closed.
