I am trying to analyze the relation on both cases in integers and positive integers, and here are what I got so far:
Both are symmetric and transitive. Because xy = yx = 1 on both cases, and if xy = 1, yz = 1, then xz = 1 also work on both cases, but I am not sure whether they're reflexive and anti-symmetric, please give me a hand, thank you
As correctly noted by the other answer, the relations in question happen to be $\{(-1,-1),(1,1)\}$ and $\{(1,1)\}$ if it were taken over $\Bbb Z$ and $\Bbb Z^+$ respectively.
These relations are indeed transitive (as any choice of $x,y,z$ (repetitions allowed) such that $x\sim y$ and $y\sim z$ result in $x\sim z$ as well) and symmetric (as any choice of $x,y$ such that $x\sim y$ results in $y\sim x$ as well), easily verified since the only examples of times where we have $x\sim y$ or similar are when $x,y$ (and $z$) are all equal to $1$ or are all equal to $-1$.
The relations are both not reflexive. This is because there exists at least one example of an element who is not related to itself. For instance, $2$ is not related to itself since $2\times 2\neq 1$.
The relations are both not irreflexive. To have been irreflexive would mean that every element is not related to itself, but this is not the case since $1$ is related to itself in both cases since $1\times 1=1$ is true.
The relations are both antisymmetric. This is because every case of $x$ being related to $y$ we have that $y$ is not related to $x$ or that $y=x$. It is indeed the case here since the only examples of times we have one thing related to another it was because they were both $1$ or were both $-1$.
I encourage you to check out my answer to Can a relation be both symmetric and antisymmetric where I give graphical interpretations of symmetry and antisymmetry described using "single-sided arrows," "doublesided arrows," and "loops." This relation is a quintessential example of one which is both symmetric and antisymmetric at the same time, the graph for it consisting only of loops.
