I will assume permutations act on the right. Throughout $i,j$ will refer to any integers with $1\le i,j\le n$.
Let $\sigma,\omega\in S_n$. Denote $\sigma^\omega=\omega^{-1}\sigma\omega$ and notice $(i^\omega)^{(\sigma^\omega)}=(i^\sigma)^\omega$ so if, for example, $\sigma=(x_1,\ldots,x_r)$ is a cycle then $\sigma^\omega=(x_1^\omega,\ldots,x_r^\omega)$.
Write $\sigma_1=(1,2,\ldots,n)$ and $\sigma_2=(1,2,\ldots, n-1)$. I will answer both interpretations of the question, so I will calculate $C_{S_n}(\sigma_1)$, $C_{S_n}(\sigma_2)$ and $C_{S_n}(\{\sigma_1,\sigma_2\})=C_{S_n}(\sigma_1)\cap C_{S_n}(\sigma_2)$.
Suppose $\omega\in C_{S_n}(\sigma_1)$, so $\sigma_1^\omega=\sigma_1$. From the above, $\sigma_1=\sigma_1^\omega=(1^\omega,\ldots,n^\omega)$. Say $1\le i\le n-1$, then $i^{\sigma_1}=i+1$ and $n^{\sigma_1}=1$. In particular if $i^\omega=j$ then (modulo $n$) $(i+1)^\omega=j+1=i^\omega+1$. Hence $\omega\in\langle \sigma_1\rangle$, but $C_{S_n}(\sigma_1)\supseteq\langle \sigma_1\rangle$ so $C_{S_n}(\sigma_1)=\langle \sigma_1\rangle$.
Suppose $\omega\in C_{S_n}(\sigma_2)$, then by the above $(n^\omega)^{\sigma_2}=(n^\omega)^{(\sigma_2^\omega)}=(n^{\sigma_2})^\omega=n^\omega$. The only fixed point of $\sigma_2$ is $n$ so $n^\omega=n$. Therefore $\omega\in S_{n-1}$ so we are reduced to the $\sigma_1$ case (but for $n-1$) so $C_{S_n}(\sigma_2)=\langle \sigma_2\rangle$.
Finally we have $C_{S_n}(\{\sigma_1,\sigma_2\})=C_{S_n}(\sigma_1)\cap C_{S_n}(\sigma_2)=\{1_{S_n}\}$
You just need to find two matrices that don't commute.
\begin{gather}
\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix}=
\begin{bmatrix}
2 & 1 \\
1 & 1
\end{bmatrix}
\\[6px]
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 \\
1 & 2
\end{bmatrix}
\end{gather}
For $n\ge2$ just take these as the upper left block and complete with ones on the diagonal and zero elsewhere. The coefficients at $(1,1)$ are different.
Best Answer
1) You are looking for a matrix $\bigl(\begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \bigr)$ which satisfies:
$\bigl(\begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \bigr)\bigl(\begin{smallmatrix} 2&0 \\ 0&3 \end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix} 2&0 \\ 0&3 \end{smallmatrix} \bigr)\bigl(\begin{smallmatrix} a&b \\ c&d \end{smallmatrix} \bigr)$
If you make the calculation you will find the following:
$2c=3c$ so that $c=0$ and $2b=3b$ so that $b=0$
Therefore the answer is $\bigl(\begin{smallmatrix} a&0 \\ 0&d \end{smallmatrix} \bigr)$ with $a \neq 0$ and $d \neq 0$ (because they are in $GL(2, \mathbb{R})$)
2) With a similar argument and calculation, you will find $\bigl(\begin{smallmatrix} a&b \\ 0&a \end{smallmatrix} \bigr)$ with $a \neq 0$