A relation for adjoint representation of $U(N)$ acting on product of matrices and $SU(2)$ generators Is the following relation true, and if so, what is the property that makes it so?
\begin{align}
    \sum_{i=1}^3\mathrm{tr}\left([U^{-1}L_iU]\phi[U^{-1}L_iU]\phi\right) \stackrel{!}{=} \sum_{i=1}^3\mathrm{tr}\left(L_i\phi L_i\phi\right)
\end{align}
where $\phi$ is an $N\times N$ Hermitian matrix, $L_i$ are the generators of $SU(2)$ in the $N$-dimensional irreducible representation, and $U \in U(N)$. For context, I arrived at this in trying to demonstrate the invariance of a more complicated expression under the $U(N)$ adjoint action appearing here: https://math.stackexchange.com/questions/4474780/invariance-of-trace-of-a-squared-commutator-under-un-conjugation-su2-repres.
 A: Here is a partial remark on a special case that might help you get started on your problem, where everything is completely calculable directly.
Take the case of N=2, where you might as well normalize your doublet generators to be the three Pauli matrices, easy to compute with. Adopt summation convention over repeated indices. Moreover, the most general 2×2 hermitean matrix can be represented as $\phi= \phi_0 {\mathbb 1}+ \phi^k \sigma^k$, so, then, note your r.h.side is an SU(2) invariant,
$$
 \mathrm{tr}\left(\sigma^i\phi \sigma^i\phi\right)=
\mathrm{tr}\Bigl(\sigma^i(\phi_0 {\mathbb 1}+ \phi^k \sigma^k) \sigma^i (\phi_0 {\mathbb 1}+ \phi^j \sigma^j)\Bigr ) \\
= 2(3\phi_0^2 -\vec \phi \cdot \vec \phi).
$$
In particular, this does not depend on the basis you use to parameterize your generators, or your vectors $\vec \phi$ in your field matrices $\phi$.
The l.h.side of your relation under scrutiny is written in the basis of rotated $\sigma^i$, namely a bispinor (adjoint) rotation $U^{-} \sigma^i U=R^i_j \sigma^j$, where the 3×3 orthogonal matrix R represents rotation by an angle double that in U, and likewise $R_k^{T~i} \phi^k$. But your expression is a scalar, and ignorant of the orientation of the axes used to express $\phi$, so the left- and right-hand sides coincide.
Embedding this argument in U(N) might take some work.
Abstractly, however, you appreciate that, in any rep, the Lie algebra elements $U^{-1} L^i U$ satisfy the very same Lie algebra of the $L^i$s, with the same structure constants, so a scalar function of the "increments" $L^i\phi$ such as the trace, would not depend on the basis.
