Center of wreath product $\mathbb{Z_n} \wr \mathbb{Z_m}$ and $\mathbb{S_m} \wr \mathbb{Z_n}$

I have a problem with calculate center of wreath products $\mathbb{Z_n} \wr \mathbb{Z_m}$ and $\mathbb{S_m} \wr \mathbb{Z_n}$. I was trying to write definitions and try do calculate something, but I don't have any idea how to do it.

Could someone help me with some ideas?

Recall if $G\curvearrowright\Omega:=\{1,\cdots,m\}$ then elements of $H\wr_\Omega G$ are of the form $(h_1,\cdots,h_m,g)$, and conjugating an element $(h_1,\cdots,h_m,e)$ by $(e,\cdots,e,g)$ yields $(h_{g^{-1}(1)},\cdots,h_{g^{-1}(m)},e)$. When multiplying two elements of $H\wr_\Omega G$, this allows us to "slide" an element of $G$ past an element of $H^\Omega$ to the right via $g\vec{h}=(g\vec{h}g^{-1})g$.

Suppose $(z_1,\cdots,z_m,\gamma)\in Z(H\wr_\Omega G)$. This is equivalent to commuting with both $H^\Omega$ and $G$, If we write out the equation $(z_1,\cdots,z_m,\gamma)(e,\cdots,e,g)=(e,\cdots,e,g)(z_1,\cdots,z_m,\gamma)$ we find $z_1=\cdots=z_m$ and $\gamma$ is central in $G$. If we write out $(z,\cdots,z,\gamma)(x_1,\cdots,x_m,e)=(x_1,\cdots,x_m,e)(z,\cdots,z,\gamma)$ we find $\gamma$ must be in the kernel $K$ of the action $G\to\mathrm{Sym}\,\Omega$.