Before I talk about $4\times 4$ matrices, let me first provide some background using the example of $2\times 2$ traceless Hermitian matrices $H$. I won't write in full rigor since I think the technical details obscure the overall idea.
Notice that $H$ is in the Lie algebra $i \mathfrak{su}(2)$ so that it can be decomposed into Pauli matrices $\sigma_1,\sigma_2,\sigma_3$, i.e., $H=c\cdot \sigma =\sum c_i \sigma_i$ where $c=(c_i)\in \mathbb{R}^3$. We also know that $H$ is unitarily diagonalizable and thus there should be some $U\in SU(2)$ which diagonalizes $H$ via the Adjoint representation, i.e., $UHU^*$ is diagonal. To find this $U$, we can use the fact that $SU(2)$ is the double covering of $SO(3)$ and the Adjoint representation of $SU(2)$ on $i\mathfrak{su}(2)$ is "equivalent" to the standard (irreducible) representation of $SO(3)$ on $\mathbb{R}^3$, that is, for every $U\in SU(2)$, there exists $R\in SO(3)$ such that $U (c\cdot\sigma)U^*=(Rc)\cdot \sigma$. Hence, for any $H$, we can always "rotate" the corresponding decomposition $c\in \mathbb{R}^3$ to point in the $x_3$-direction, i.e., $UHU^* = |c| \sigma_3$, so that $H$ is now diagonalized.
Now, let me consider $4\times4$ traceless Hermitian matrices $H$. Indeed, $H$ is in the 15-dim Lie algebra $i\mathfrak{su}(4)$, which is isomorphic to $\mathfrak{so}(6)$ since $SU(4)$ is the double cover of $SO(6)$. Now, if I want to diagonalize $H$ via some $U\in SU(4)$, you would think that I could do something similar. However, I can't quite think of a way to do it, since the Adjoint representation of $SU(4)$ is definitely not equivalent to the standard $SO(6)$ representation on $\mathbb{R}^6$ (dimensions are not the same). Essentially, what I think I'm asking is what is the "equivalent" representation of the Adjoint representation of $SU(4)$ if we were to think in terms of $SO(6)$. I'm currently thinking that it might be related to the induced action of $SO(6)$ on the anti-symmetric tensor product $\wedge^2 \mathbb{R}^6$, i.e., $R\cdot (x\wedge y) = Rx \wedge Ry$ since it has the right number of dimensions, but I'm not quite sure of the details.
EDIT. @Callum Thanks for the hint. Indeed, $\mathfrak{so}(6)$ is the real vector space of anti-symmetric matrices, so that isomorphism $|a\rangle \langle b|-|b\rangle \langle a| \mapsto a\wedge b$ makes sense (physicist notation). Hence, the induced action of $SO(6)$ on $\wedge^2 \mathbb{R}^6$, i.e., $R\cdot(x\wedge y)= Rx\wedge Ry$ is just the Adjoint action on $\mathfrak{so}(6)$. So, now all I really need to understand that is the explicit Lie algebra isomorphism $\mathfrak{su}(4) \mapsto \mathfrak{so}(6)$ to obtain the diagonalization, which seems to be answered by this question. Ultimately, it seems that diagonalization is not that straightforward as the $2\times 2$-matrix, mainly due to the fact that $H$ can be expressed as a nontrivial sum of anti-symmetric tensors, e.g., $x\wedge y + z\wedge w +\cdots$, and simultaneously rotating all the vectors $x,y,z,w,...$ to the right axes is quite difficult to imagine.
