If all the diagonal entries of$\Lambda$ are distinct, it commutes only with diagonal matrices.
In contrast, for each $k$ consecutive equal diagonal entries in $\Lambda,$ we may allow $A$ to have anything at all in the corresponding $k$ by $k$ square block with both corners on the main diagonal.
This means that the set of matrices that commute with $\Lambda$ has a minimum dimension $n$ and a maximum dimension $n^2.$ Suppose we have $r$ different diagonal entries, and there are $k_i$ copies of diagonal entry $\lambda_i.$ Each $k_i \geq 1,$ and we have
$$ k_1 + k_2 + \cdots + k_r = n. $$
Then by the block construction I mentioned above, the dimension of the space of matrices that commute with $\Lambda$ is
$$ k_1^2 + k_2^2 + \cdots + k_r^2. $$
The minimum is when $r=n,$ so all $k_i = 1,$ and the dimension is $n$
The maximum is when $r=1,$ and $k_1=n,$ the matrix is a scalar multiple of the identity matrix, and the dimension is $n^2.$