I have come up with a line of reasoning that appears to constrain the spectrum of a generic Hermitian matrix to consist of integer multiples of $2\pi$. Of course this is nonsense, but I cannot find the point at which the argument fails. Any help would be greatly appreciated.
Let $H$ be a generic $N\times N$ Hermitian matrix, and let $G$ be the maximal group of unitary matrices that commute with $H$, $G=\{U\in\mathsf{U}(N)\, |\, UHU^\dagger=H\}$. The matrices $U$ form a representation of $G$ (by definition), and since this representation is unitary and finite-dimensional it is fully reducible. Therefore there exists some unitary basis transformation $V$ such that $$ \tilde{U}(U) := VUV^\dagger=R_1(U)^{\oplus n_1}\oplus\cdots\oplus R_k(U)^{\oplus n_k}=\left[\mathbb{I}_{n_1}\otimes R_1(U)\right]\oplus\cdots\oplus\left[\mathbb{I}_{n_k}\otimes R_k(U)\right] $$ where the $R_i$ are irreducible and mutually inequivalent representations of $G$. The dimension of $R_i$ is $d_i$, and the multiplicity of $R_i$ in the defining representation of $G$ is $n_i$; these satisfy $\sum_{i=1}^k n_i d_i=N$. Since $\tilde{H}=VHV^\dagger$ commutes with all the $\tilde{U}$, it follows from Schur's lemma that this matrix must have the form $$ \tilde{H} = \left[h_1\otimes \mathbb{I}_{d_1}\right]\oplus\cdots\oplus\left[h_k\otimes \mathbb{I}_{d_k}\right], $$ where $h_i$ is a $n_i\times n_i$ Hermitian matrix.
So far this is all quite clear-cut and straightforward. What confuses me is the following:
Suppose that there exists a matrix $u_1\in \mathsf{U}(n_1)$ that commutes with the matrix $h_1$. Then the unitary matrix $$ \tilde{U}_1 = \left[u_1\otimes\mathbb{I}_{d_1}\right]\oplus\left[\mathbb{I}_{n_2}\otimes\mathbb{I}_{d_2}\right]\oplus\cdots\oplus\left[\mathbb{I}_{n_k}\otimes\mathbb{I}_{d_k}\right] $$ commutes with $\tilde{H}$ so, since $G$ is maximal, there must be some $U$ for which $\tilde{U}_1$ is equal to $\tilde{U}(U)$ (up to a possible phase). Equating $\tilde{U}_1=\tilde{U}(U)$ implies that $R_i(U)=\mathbb{I}_{d_i}, i=2, \ldots, k$ and so $\tilde{U}_1=\tilde{U}(\mathbb{I}_N)$. This implies that the matrix $u_1=\mathbb{I}_{n_1}$ (again, up to a phase), so the only unitary matrix that commutes with $h_1$ is a multiple of the identity. Now, since $h_1$ is Hermitian, it commutes with the unitary matrix $\exp(\mathrm{i}\theta h_1)$ for arbitrary $\theta\in\mathbb{R}$; it then follows from the preceding argument that $\exp(\mathrm{i}\theta h_1)=\exp(\mathrm{i}\phi) \mathbb{I}_{n_1}$ for some $\phi\in\mathbb{R}$. It then follows that $\theta h_1$ is similar to a diagonal matrix of the form $\mathrm{diag}(2\pi a_1, \ldots, 2\pi a_{n_1})$ with $a_i\in\mathbb{Z}$. Repeating this argument for all $h_i$ appears to show that the spectrum of $H$ consists only of integer multiples of $2\pi$, but of course this is nonsense as $H$ was chosen arbitrarily. At what point(s) does the above argument fail? Moreover, what can one say about the matrices $h_i$ and $u_i$ in general?
The line
is stated with no justification and I can find no justification, and of course it is obviously false if you work through an actual example. $\theta$ was arbitrary but somehow it disappeared so something very fishy has happened in this step. A generic Hermitian matrix $H$ diagonalizes to a diagonal matrix with distinct real entries so each of the $h_i$ is just a real scalar given by an eigenvalue of $H$.