You may directly compute the conjugates of each element
but we are going to use the following theorem to simplify the computations.

Theorem.

The number of conjugates of an element $g$ in a group is the index $|G: C_G(s)|$ of the centralizer of $s$.

Solution.

Let us denote $G=D_8$.
Let $K_x$ be the conjugacy class in $G$ containing the element $x$.

Note that $\langle r \rangle < C_G(r) \lneq G$ and the order $|\langle r \rangle|=4$.
Hence we must have $C_G(r)=\langle r \rangle$.
Thus the element $r$ has $|G:C_G(r)|=2$ conjugates in $G$.

Since $srs^{-1}=r^3$, the conjugacy class $K_r$ containing $r$ is $\{r, r^3\}$.

Since $\langle s \rangle < C_G(s) \lneq G$ and $|\langle s \rangle|=2$, we have either $C_G(s)=\langle s\rangle$ or $|C_G(s)|=4$.
Since $r^2s=sr^2$, we must have $|C_G(s)|=4$ and hence the conjugacy class $K_s$ containing $s$ has $|G:C_G(s)|=2$ elements.

Centralizer, Normalizer, and Center of the Dihedral Group $D_{8}$
Let $D_8$ be the dihedral group of order $8$.
Using the generators and relations, we have
\[D_{8}=\langle r,s \mid r^4=s^2=1, sr=r^{-1}s\rangle.\]
(a) Let $A$ be the subgroup of $D_8$ generated by $r$, that is, $A=\{1,r,r^2,r^3\}$.
Prove that the centralizer […]

The Order of a Conjugacy Class Divides the Order of the Group
Let $G$ be a finite group.
The centralizer of an element $a$ of $G$ is defined to be
\[C_G(a)=\{g\in G \mid ga=ag\}.\]
A conjugacy class is a set of the form
\[\Cl(a)=\{bab^{-1} \mid b\in G\}\]
for some $a\in G$.
(a) Prove that the centralizer of an element of $a$ […]

The Center of a p-Group is Not Trivial
Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$.
(Such a group is called a $p$-group.)
Show that the center $Z(G)$ of the group $G$ is not trivial.
Hint.
Use the class equation.
Proof.
If $G=Z(G)$, then the statement is true. So suppose that $G\neq […]

Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set
Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$.
For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$.
Proof.
$(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$.
Take any $h\in C_G(X)$. Then for […]

Normalizer and Centralizer of a Subgroup of Order 2
Let $H$ be a subgroup of order $2$. Let $N_G(H)$ be the normalizer of $H$ in $G$ and $C_G(H)$ be the centralizer of $H$ in $G$.
(a) Show that $N_G(H)=C_G(H)$.
(b) If $H$ is a normal subgroup of $G$, then show that $H$ is a subgroup of the center $Z(G)$ of […]

Dihedral Group and Rotation of the Plane
Let $n$ be a positive integer. Let $D_{2n}$ be the dihedral group of order $2n$. Using the generators and the relations, the dihedral group $D_{2n}$ is given by
\[D_{2n}=\langle r,s \mid r^n=s^2=1, sr=r^{-1}s\rangle.\]
Put $\theta=2 \pi/n$.
(a) Prove that the matrix […]

The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]

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