Basically the title.
For $A \in \mathbb{R}^{n \times n}$, how does $\text{rank}(A)$ relate to $\text{rank}(\text{Diag}(\text{Vec}(A)))$ where $\text{Diag}(\text{Vec}(A) \in \mathbb{R}^{n^2 \times n^2}$? Here, $\text{Vec}$ is the vectorization operator, and for $x$ a column vector, $\text{Diag}(x)$ is the matrix with $x$ as its diagonal and $0$ everywhere else.
I have tried writing $\text{Diag}(\text{Vec}(A))$ as
$$\text{Diag}(\text{Vec}(I_n))(I_n \otimes A)\text{Vec}(I_n)1_{n}^{T},$$
where $1_n^{T}$ is a $n$ long vector of $1's$ (this was obtained using standard matrix theory), but I couldn't proceed from here.
I suspect that $\text{rank}(A)=\text{rank}(\text{Diag}(\text{Vec}(A)))$ but I do not know how to prove it!