Proofs on relations, equivalence relations, and inverses Suppose that $X$ is a nonempty set. Let $A$ denote the set of functions
from $X$ to $X$. Let $B$ denote the set of $h \in A$ such that $h$ is bijective.
Define a relation $R$ on $A$ by
$$R = \{(f, g) \in A \times A : ∃h \in B \text{ such that }h \circ f = g \circ h\}.$$
Observe that if 
$$(f, g) ∈ A × A, h \in B$$
and 
$$h \circ f = g \circ h$$
then 
$$h^{-1} \circ h \circ f = h^{−1}\circ g \circ h.$$


*

*Prove that $R$ is an equivalence relation on $A$.

*Prove that if $(f, g) \in R$, then $(f \circ f, g \circ g) \in R$.

*Suppose that $(f, g) \in R$. Suppose that there is some $x \in X$ with
$f(x) = x$. Prove that there is some $y \in X$ with $g(y) = y$.


I know what an equivalence relation is, but I have no clue how to prove it effectively. I also have no clue where to start on the other two questions or what the prompt is actually saying. Any help will be greatly appreciated.
 A: I'll use Greek letters for members of $B$.
I'll also drop the use of $\circ$, and concatenate function symbols.
Additionally, I'll use $fRg$ for $(f,g) \in R$.
For (1), you have to prove that $R$ is reflexive, that is $fRf$, for which you can just take $\iota_X$, the identity of $X$, which satisfies $\iota_Xf = f\iota_X$, for all $f \in A$.  
Then you must prove it is symmetric, that is, $fRg$ implies $gRf$.
Now, if $fRg$ then $\alpha f = g\alpha$, for some $\alpha \in B$.
It follows that $\alpha^{-1}\in B$ and
$$\alpha^{-1}g = \alpha^{-1}g\alpha\alpha^{-1} = \alpha^{-1}\alpha f\alpha^{-1} = f\alpha^{-1}.$$
Finally, you have to prove that $R$ is transitive, that is, if $fRg$ and $gRh$ then $fRh$.
So suppose that $fRg$ and $gRh$ and for $\alpha, \beta \in B$, we have
$\alpha f = g\alpha$  and $\beta g = h \beta$.
Check that $\beta\alpha f = h\beta\alpha$.
Thus, $R$ is indeed, an equivalence relation.
(2) is straightforward: if $\alpha f = g\alpha$, then
$$\alpha f^2 = (\alpha f) f = (g\alpha) f = g(\alpha f) = g (g\alpha) = g^2\alpha.$$
For (3), if $f(x_0)=x_0$ and $\alpha \in B$ is such that $\alpha f = g\alpha$, then take $y_0 = \alpha(x_0)$, and
$$y_0 = \alpha(x_0) = \alpha f(x_0) = g\alpha(x_0) = g(y_0).$$
