I’m afraid that none of it makes any sense. If $R$ is an equivalence relation on $A$, $R$ is a subset of $A\times A$, so it can’t possibly be an element of $A$. Let’s look at a very small example. Suppose that $A = \{0,1\}$ and $R = \{(0,0),(1,1)\}$; this $R$ really is an equivalence relation on this $A$, so the example is certainly relevant. (In fact, it’s the relation of equality on $A$.) The elements of $A$ are $0$ and $1$; does $\{(0,0),(1,1)\}$ look like either of them?
Your hypothesis is that $R$ is an equivalence relation on $A$. This means, first of all, that it’s a relation on $A$, i.e., a subset of $A\times A$: every member of $R$ is an ordered pair of the form $(a,b)$, where $a$ and $b$ are members of $A$. But it’s not just any old relation on $A$: it’s reflexive, symmetric, and transitive. The first of these means that for every $a \in A$, $(a,a)\in R$. The second means that if $(a,b)\in R$, then $(b,a)\in R$. And the third means that if $(a,b)\in R$ and $(b,c)\in R$, then $(a,c)\in R$.
You want to show that $R^{-1}$ is also an equivalence relation on $A$. So what is $R^{-1}$? By definition $R^{-1} = \{(b,a):(a,b)\in R\}.$ (Note that we’re talking about ordered pairs in $R$, not in $A$: they aren’t in $A$.) To show that $R^{-1}$ is an equivalence relation on $A$, you have to show
$\qquad$(1) that $R^{-1}$ is a relation on $A$;
$\qquad$(2) that $R^{-1}$ is reflexive;
$\qquad$(3) that $R^{-1}$ is symmetric; and
$\qquad$(4) that $R^{-1}$ is transitive.
In other words, you have to show
$\qquad$(1) that $R^{-1}\subseteq A\times A$;
$\qquad$(2) that for every $a\in A$, $(a,a)\in R^{-1}$;
$\qquad$(3) that if $(a,b)\in R^{-1}$, then $(b,a)\in R^{-1}$; and
$\qquad$(4) that if $(a,b)\in R^{-1}$ and $(b,c)\in R^{-1}$, then $(a,c)\in R^{-1}$.
In order to do this, you’ll have to use the facts about $R$ that I mentioned in the first paragraph.