How to solve cycle equations in Group Theory? Is my approach correct?

Question

I am studying Abstract Algebra on my own so I don't know if there are gaps in my knowledge. I didn't find any example to solve the following equation in my reference textbook.
Find ${\tau\in S_7, \text{such that } \sigma'=\tau\sigma\tau^{-1} \text{ where } \sigma=(125)(3674), \sigma'=(314)(2765)}$
I tried the following.
Right multiplication by ${\tau}$ and left multiplication by ${\sigma'^{-1}}$ gives ${\tau=\sigma'^{-1}\tau\sigma}$
${\implies \tau=(341)(2765)^3\tau(125)(3674)}$
I tried the following approach.
$\tau(1)=(341)(2765)^3\tau(\sigma(1))=(341)(2765^3)\tau(2)$
$\tau(2)=(341)(2765)^3\tau(\sigma(2))=(341)(2765)^3\tau(5)$
$\tau(5)=(341)(2765)^3\tau(\sigma(5))=(341)(2765)^3\tau(1)$
If I back substitute the values, I get
$\tau(2)=(341)^2(2765)^6\tau(1) \therefore\tau(1)=(2765)\tau(1)$
Is this approach correct? Am I close? How do I proceed forward?

One observation I made was if $\tau(1)\in\{2,5,6,7\}, $ then the equation cannot be satisfied. Therefore, $\tau(1)\in\{3,4,1\}$. Is this correct?

Answer 1

Hint: Think of conjugation as a "change of basis".

So $(125) $ must be mapped to $(314)$.

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In fact, for every "change of basis", we have such a conjugation.
What is the conjugation that maps 1 to 3, 2 to 1, 5 to 4, 3 to 2, 6 to 7, 7 to 6 and 4 to 5?

$ 1 \rightarrow 3 \rightarrow 2 \rightarrow 1, 4 \rightarrow 5 \rightarrow 4, 6 \rightarrow 7 \rightarrow 6$.
Hence, this is $ (132) (45)(67)$.

What is we wanted to mix it up so that we map 1 to 1, 2 to 4, 5 to 3, 3 to 6, 6 to 5, 7 to 2, 4 to 7?

$ 1 \rightarrow 1, 2 \rightarrow 4 \rightarrow 7 \rightarrow 2, 3 \rightarrow 6 \rightarrow 5 \rightarrow 3$.
Hence, this is $ (1)(247)(365)$.