I have an operator H and I want to perform the transformation $$ e^{i S}H e^{-iS} $$ where $S$ is a diagonal operator with dimension $L$, for instance $$S=\text{diag}\left(s_1,\ldots,s_L\right)$$ where $s_j$ are distinct entries, and $H$ a generic symmetric $L\times L$ operator. By definition of the matrix exponential one can write $$ e^{i S}H e^{-iS}=\sum_{n,m=0}^{\infty}\frac{(-1)^m (i)^{n+m}}{n! m!} S^n H S^m $$ or also $$ e^{i S}H e^{-iS}=\sum_{m=0}^{\infty}\frac{1}{m!}[iS,H]_m$$, where $[iS,H]\equiv [iS,[iS,\ldots,[iS,H]]]$ is the nested commutator.
Using the propriety that $S$ is diagonal it is possible to find a general result of this operation, considering that the commutator $[S,H]$ is known?
Exponentiating a diagonal matrix is nice: $[e^{D}]_{jk} = e^{[D]_{jk}} \delta_{jk}$ whenever $D$ is diagonal. So $$[e^{iS} H e^{-iS}]_{jk} = \sum_{ \ell ,m} [e^{iS}]_{j \ell} [H]_{\ell m} [e^{-iS}]_{mk} =\sum_{ \ell ,m} e^{i[S]_{j \ell}} [H]_{\ell m} e^{-i[S]_{mk}}\delta_{j \ell} \delta_{mk} = e^{i(s_j-s_k)} [H]_{j k}. $$ The transformation results in the $jk^{th}$ element of $H$ being multiplied by the phase $e^{i(s_j-s_k)}$. If every $s_j$ is real, this is an example of a unitary transformation, as $$e^{iS} (e^{iS})^*= e^{iS}e^{-iS} = I,$$ the identity matrix. In particular, if the matrix $S$ is diagonal in the basis $\{\mathbf e_1,\mathbf e_2,...,\mathbf e_L\},$ the transformation is equivalent to a change of basis $ \mathbf e_j \mapsto e^{is_j} \mathbf e_j$.