Let $\tau = (1\,2\,3)(4\,5\,6)(7\,8\,9)(10\,11\,12) \in S_{12}$. We want to find a permuation $\sigma$ of order 4 such that $\sigma^{-1}\tau\sigma = \tau^2$.
By brute forcing with cases, I arrived at one possible $\sigma = (1\,6\,7\,12)(2\,5\,8\,11)(3\,4\,9\,10)$, but it wasn't elegant by any means. I was wondering what might be a systematic way of solving these types of problems. Any input would be appreciated!
I will assume that your permutations are acting on the right - you may have to swap some orders and interchange $\sigma$ with $\sigma^{-1}$ if this is not the case.
You have $\tau^2=(1,3,2)(4,6,5)(7,9,8)(10,12,11)$.
In general, you can show that $\sigma^{-1}(x_1,\ldots, x_k)\sigma=(x_1^\sigma,\ldots,x_k^\sigma)$.
With these two things in mind, you can read off a choice of $\sigma$: $$\sigma=(2,3)(5,6)(8,9)(11,12)$$
Of course this has order $2$, not $4$. One way to resolve this is to find an order $4$ permutation $\sigma_0$ which commutes with $\sigma$ and $\tau$ and use $\sigma_1=\sigma_0\sigma$. This would give $\sigma_1^{-1}\tau\sigma_1=\tau^2$ and the order of $\sigma_1$ is $4$. One choice for $\sigma_0$ and corresponding $\sigma_1$ is: $$\sigma_0=(1,4,7,10)(2,5,8,11)(3,6,9,12)$$ $$\sigma_1=(1,4,7,10)(2,6,8,12)(3,5,9,11)$$