Let $H$ be a finite dimensional Hermitian operator in $\mathfrak{gl}(n;\mathbb{C})$ with adjoint representation $\text{ad}_H$. Using the argument presented below, I end up with the result that $\text{ad}_H = 0$ using only the assumption that $H = H^\dagger$, and I am wondering where my error is, given that this is not necessarily true.
Since $H$ is Hermitian, its eigenvalues $\{\lambda_i\}_{i=1}^{n}$ are all real. Since the eigenvalues of the adjoint representation are the set of differences $\{\lambda_i - \lambda_j: 1 \leq i,j\leq n\}$ of the eigenvalues of $H$, it follows that the adjoint representation has only real eigenvalues, and is thus Hermitian: $$\text{ad}_H = (\text{ad}_H)^\dagger$$
Second, noting that, for $X \in \mathfrak{gl}(n;\mathbb{C})$, we have $[X^\dagger, \cdot] = -[\cdot, X^\dagger] = -[X,\cdot]^\dagger,$ it follows that for arbitrary $X$, we have $$\text{ad}_{X^\dagger} = -(\text{ad}_X)^\dagger.$$
Combining the two results above for $H$, we have
$$\text{ad}_H = (\text{ad}_H)^\dagger \quad \text{and} \quad \text{ad}_H = \text{ad}_{H^\dagger} = -(\text{ad}_H)^\dagger,$$
meaning that $\text{ad}_H$ is both Hermitian and anti-Hermitian, and thus, is equal to $0$. This is clearly not necessarily true, so could someone help me identify where the error in my argument is?
In your second bullet point, the second equality seems to be false.
The correct identity is
$$-[\cdot,X^\dagger]=-[X,\cdot^\dagger]^\dagger.$$