Permutations: A cycle is conjugate to its own inverse I need help with d) here.

Let $2 \le r \le n$ be two natural numbers. Assume that $\rho \in S_n$
  is a permutation of the set $I_n=\{1,2,...,n\}$. Let $x_i \in I_n$ for
  $1 \le i\le r$ be $r$ different numbers.
a) Show that
  $\rho(x_1,...,x_i,...,x_r)\rho^{-1}=(\rho(x_1),...,\rho(x_i),\rho(x_{i+1}),...,\rho(x_r))$.
  HINT: For $i < r$ feed the left side with $\rho(x_i)$ and check that
  you get $\rho(x_{i+1})$.
b) Show that: $(2, 4)(1 ,5)(1, 2 ,3 ,4, 5)(2, 4)(1, 5)=(5,4,3,2,1)$.
c) Let $\sigma$ be the $r$-cycle $(1,2,3,...,r)$ in $S_n$. Show that
  $\sigma$ is conjugate to its own inverse; that is, there is a
  permutation such that $\rho\sigma\rho^{-1}=\sigma^{-1}$.
d) Show that in c) one may take for $\rho$ a permutation that fixes
  any of the numbers that $\sigma$ moves. (This means: Pick one
  $1 \le i\le r$, then one may find a $\rho$ with $\rho(i)=i$).

If $r$ is odd I can just generalize b), but what if $r$ is even? I don't see how to solve the exercise then.
 A: To answer the d) question. Fix $1\leq i\leq r$ such an integer then you can write :
$$\sigma:=(1,2,...,r)=(i,i+1,...,r,1,...,i-1)$$
This is just another way of writing your permutation.
Now you know that :
$$\sigma^{-1}=(1,r,r-1,...,2)=(i,i-1,...,1,r,...,i+1) $$
And you want $\rho$ such that :
$$\rho\sigma\rho^{-1}=\sigma^{-1} \text{ and } \rho(i)=i$$
From a) you get :
$$\rho\sigma\rho^{-1}=(\rho(i),\rho(i-1),...,\rho(1),\rho(r),...,\rho(i+1))=(i,\rho(i-1),...,\rho(1),\rho(r),...,\rho(i+1))$$
Finally you just want :
$$(i,\rho(i+1),...,\rho(r),\rho(1),...,\rho(i-1))=(i,i-1,...,1,r,...,i+1) $$
You can now see that such a $\rho$ always exists. You just want to send the $r-1$ tuple:
$$(i+1,...,r,1,...,i-1)\text{ on } (i-1,...,1,r,...,i+1)$$
with a permutation of the set $\{1,...,i-1,i+1,...,r\}$ and this can be done because the action of the group permutation is $r-1$ transitive. 
A: Hint: consider the permutation $(1,2,3,4)$. What can you say of the permutation $(1,4,3,2)$?

Suppose you want a $\rho$ that fixes $i$. Since you can always write $\sigma^{-1}$ with $i$ as the leading number, without loss of generality we may assume that $i = 1$. Now, by a) you know that
$$
\rho \sigma \rho^{-1} = (\rho(1),\rho(2),\dotsc,\rho(r))
$$
so it is enough to choose a $\rho$ that sends the sequence $2,3,\dotsc,r$ into the sequence $r,r-1,\dotsc,2$ and fixes every other number in $\{1,\dotsc,n\}$.
