Find all permutations such that $\sigma=a\tau a^{-1}$

Question

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For (b) and (c) we note that $\sigma$ and $\tau$ have different parity so there cannot be any $a\in S_4$ that will fix that parity mismatch.

For (a) we have the cycle $a^{-1}=(3 2 4)$ and it is unique. But why is it unique because it needs to be EVEN so it can only be a 3-cycle.

Does this make sense?

EDIT:

$(13)(24)$ works as well for (a) so now I am not sure how to find a conclusive answer.

Answer 1

Consider transpositions.

Number 4 is a fixed point of $\tau$, while number 2 is a fixed point of $\sigma$. So you must send 4 into 2. This gives the first transposition, (24).

Now this would send $\tau$ into $(143)(2)$, which is not correct. How can you change the first cycle into the desired $(134)$? There are 3 possibilities. You can change (14), or (13) or (34).

These possibilities give $a=(14)(24)=(142)$, $a=(13)(24)$ and $a=(34)(24)=(234)$.

That's all