I am studying graded vector spaces and I have a simple question. Let me denote by $\mathbb Z_2=\mathbb Z\text{mod}2=\{-1,0,+1\}$. Now let's perform a shift by an integer $k$ and get the set $\mathbb Z_2+k=\mathbb Z\text{mod}2+k=\{k-1,k,k+1\}$.
Intuitively, I have the feeling that the two sets $\mathbb Z_2$ and $\mathbb Z_2+k$ are equivalent, in that a $\mathbb Z_2$-graded vector space has the same decomposition in vector subspaces as a $\mathbb (\mathbb Z_2+k)$-graded vector space. The only difference is the degrees of the homogeneous elements in each space, which are naturally shifted by $k$.
Is the $(\mathbb Z_2+k)$-graded vector space the same as the initial $\mathbb Z_2$-graded vector space? If yes, is it because of some sort of isomorphism between the two sets $\mathbb Z_2$ and $\mathbb Z_2+k$?