Simultaneously Diagonalizable

Statement Ⅰ is true since diagonal matrices commute.

Statement Ⅱ is true: Perform a similarity transformation on both $A$ and $B$ that diagonalizes $A$ (but not necessarily $B$ ) and group together any repeated eigenvalues of $A.$ If $\mu_1,\mu_2,\dots,\mu_d$ are the distinct eigenvalues of $A$ and $n_1,n_2,\dots,n_d$ are their respective multiplicities, then we may assume that $A= \begin{bmatrix} \mu_1I_{n_1} & & & 0 \\ & \mu_2I_{n_2} & & \\   & & \ddots  &  \\   0 & & & \mu_dI_{n_d} \\ \end{bmatrix},\quad \mu_i\neq\mu_j,\text{ if }i\neq j$ Since $AB=BA,$ we can see, by multiplication of block matrix, that $B= \begin{bmatrix} B_1 & & & 0 \\ & B_2 & & \\   & & \ddots  &   \\   0 & & & B_d \\ \end{bmatrix},\quad \text{each }B_i\in M_{n_i}(\mathbb C)$ is block diagonal conformal to $A.$ Since $B$ is diagonalizable, we can see, by induction, that each $B_i$ is diagonalizable. Let $T_i\in M_{n_i}(\mathbb C)$ be nonsingular and such that $T_i^{-1}B_iT_i$ is diagonal for each $i=1,2,\dots d.$ Let $T= \begin{bmatrix} T_1 & & & 0 \\ & T_2 & & \\   & & \ddots  &   \\   0 & & & T_d \\ \end{bmatrix}$ Then $T_i^{-1}\mu_iI_{n_i}T_i=\mu_iI_{n_i},$ so $T^{-1}AT=A$ and $T^{-1}BT$ are both diagonal.

---

Generalization for any number of matrices could be:

Let $\mathcal F\subset M_n(\mathbb C)$ be a family of diagonalizable matrices. Then $\mathcal F$ is a commuting family , that is, every pair of matrices in $\mathcal F$ commute, if and only if it is a simultaneously diagonalizable family , that is, there is a single nonsingular $S\in M_n(\mathbb C)$ such that $S^{-1}AS$ is diagonal for every $A\in\mathcal F.$

$$

References

[1] Roger A. Horn and Charles R. Johnson, Matrix analysis , Cambridge University Press, 2013.