What is the inverse cycle of permutation?

To elaborate on my comment:

To find the inverse of a permutation just write it backwards. If $\tau = (1243)(67)$ then $\tau^{-1}=(76)(3421)$ which can then be rewritten as $\tau^{-1}=(1342)(67)$.

For for the above question: $(123)^{-1}=(321)=(132)$.

How does one prove this?

First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$.
Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_k \mapsto a_1$,
then $\sigma^{-1}$: $a_k \mapsto a_{k-1} \mapsto a_{k-2} \mapsto \cdots \mapsto a_1 \mapsto a_k$.
This is precisely the cycle $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards.

Now what about a list of cycles? Say $\sigma=\sigma_1\cdots \sigma_\ell$. Recall that $\sigma^{-1}=(\sigma_1\cdots \sigma_\ell)^{-1}=\sigma_\ell^{-1}\cdots \sigma_1^{-1}$. So we reverse the list of cycles and then write each one backwards -- thus the inverse is just the whole thing written backwards.

One thing to note: This still works even if $\sigma$ is not written in terms of disjoint cycles.

So, you have that $\sigma_1$ is the cycle such that,

$$\begin{align} 1 \mapsto 2 \\ 2 \mapsto 3 \\ 3 \mapsto 1\end{align}$$

It's inverse, $\sigma_1^{-1}$ is a cycle such that composition, $ \sigma_1 \circ \sigma_1^{-1}=\sigma_1^{-1} \circ \sigma_1$ is identity. So, the inverse cycle should look like :

$$\begin{align} 2 \mapsto 1 \\ 3 \mapsto 2 \\ 1 \mapsto 3\end{align}$$

What is this in cycle notation?

$\sigma_1^{-1} \equiv(213)$

I'll let you try the other one.

A particularly easy way of doing this, once you understand what the inverses do is: just to write the cycle backwards!

Note that for $(123)$, this is just $(321)$. Now, recall, that set of all permutations form a group. In a group, inverses are unique. So, can you tell me why $(321)$ and $(213)$ are the same?

Another way to look at Bill Cook's fifth paragraph in his answer:

First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$. Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_{k - 1} \mapsto a_k \mapsto a_1$, then $\sigma^{-1}$:
$a_1 \color{aqua}{\leftarrow} a_2 \color{aqua}{\leftarrow} a_3 \color{aqua}{\leftarrow}\cdots \color{aqua}{\leftarrow} a_{k - 1} \color{aqua}{\leftarrow} a_k \color{aqua}{\leftarrow} a_1 \iff$
$a_k \color{aqua}{\mapsto} a_{k-1} \color{aqua}{\mapsto} a_{k-2} \color{aqua}{\mapsto} \cdots \color{aqua}{\leftarrow} a_2 \color{aqua}{\mapsto} a_1 \color{aqua}{\mapsto} a_k$.
This is precisely the cycle $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards.