Permutation and Equivalence Let X be a nonempty set and define the two place relation ~ as 
$\sigma\sim\tau$ if and only if $\rho^{-1}\circ\sigma\circ\rho=\tau$ for some permutation $\rho$ 
For reflexivity this is what I have: 
Let x$\in$X such that $(\rho^{-1}\circ\sigma\circ\rho)(x)=\sigma(x)$ 
Than $\rho^{-1}(\sigma(\rho(x)))=\sigma(x)$
So $\sigma(\rho(x))=\rho(\sigma(x))$
Therefore $\sigma(x)=\rho^{-1}(\rho(\sigma(x))$
So $\sigma(x)=\rho^{-1}(\sigma(\rho(x))$
Finally $\sigma(x)=\rho^{-1}\circ\sigma\circ\rho$ 
Does that show that the relation is reflexive? 
 A: You assumed, I believe, what you were to prove. Look at your expression following "Let" and look at your expression following "Finally"...they say the same thing!
How about letting $\rho = \sigma$: all that matters to show is that for each $\sigma \in X$, $\sigma \sim \sigma$, there exists some permutation in $X$ satisfying the relation: 
This reflexive relation is satisfied for any $\sigma \in X$ by choosing $\rho$ to be itself: for $\tau$, let $\rho = \tau$...etc...
Then $$(\sigma^{-1} \circ \sigma \circ \sigma)(x) = (\sigma^{-1} \circ \sigma)(x) \circ \sigma(x) $$ $$\vdots$$
$$ = \sigma(x)$$.
A: It does not, I’m afraid. You’ve apparently misunderstood what you have to prove in order to show that $\sim$ is reflexive. You must show that if $\sigma$ is any permutation of $X$, then $\sigma\sim\sigma$, which means that there is some permutation $\rho$ of $X$ such that $\rho^{-1}\circ\sigma\circ\rho=\sigma$. This in turn means that for each $x\in X$, $(\rho^{-1}\circ\sigma\circ\rho)(x)=\sigma(x)$, i.e., that for each $x\in X$, 
$$\rho^{-1}\left(\sigma\big(\rho(x)\big)\right)=\sigma(x)\;.$$
What if $\rho$ did nothing at all to each element of $X$, so that $\rho(x)=x$ for each $x\in X$? Then you’d have
$$\rho^{-1}\left(\sigma\big(\rho(x)\big)\right)=\rho^{-1}\big(\sigma(x)\big)\;.$$
And since $\rho$ does nothing, $\rho^{-1}$, which undoes $\rho$, must also do nothing, and we have
$$\rho^{-1}\left(\sigma\big(\rho(x)\big)\right)=\rho^{-1}\big(\sigma(x)\big)=\sigma(x)\;.$$
It works! If we take $\rho$ to be the identity permutation, sometimes written $1_X$, we have $1_X^{-1}\circ\sigma\circ 1_X=\sigma$, no matter which permutation of $X$ $\sigma$ might be. This shows that $\sigma\sim\sigma$ for each permutation $\sigma$ of $X$ and hence that $\sim$ is reflexive.
