Suppose I have a square matrix over the complex numbers $A$ of dimension $n$, and define $X_3(A) = R_3 \otimes A + {R_3}^{\dagger} \otimes A^{\dagger}$ where $R_3 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ (where the $0$ entries are the $O_n$ matrix), so that $X_3(A) = \begin{bmatrix} 0 & 0 & A \\ 0 & 0 & 0 \\ A^\dagger & 0 & 0\end{bmatrix}$.
Let $U_1 = \begin{bmatrix} \sqrt{-i} I & 0 & 0 \\ 0 & I & 0 \\ 0 & 0 & \sqrt{i} I\end{bmatrix}$, which is unitary,
and I have written in my notes for a course that $e^{iX_3(A)} = U_1 e^{iX_3(iA)}U_1^{\dagger}$. I am struggling to understand why this is true. I understand that if a matrix $M = U D U^{-1}$, and $D$ is diagonal, then $e^M = Ue^D U^{-1}$, but in this case $e^{iX_3(iA)}$ is not diagonal so I am not sure how this result was obtained.
This has nothing to do with diagonal matrices. In general we have \begin{aligned} U\exp(M)U^{-1} &=U\left(I+M+\frac1{2!}M^2+\frac1{3!}M^3+\cdots\right)U^{-1}\\ &=I+UMU^{-1}+\frac1{2!}UM^2U^{-1}+\frac1{3!}UM^3U^{-1}+\cdots\\ &=I+UMU^{-1}+\frac1{2!}(UMU^{-1})^2+\frac1{3!}(UMU^{-1})^3+\cdots\\ &=\exp(UMU^{-1}). \end{aligned}