Find all permutations $\alpha \in S_7$ such that $\alpha^3 = (1 2 3 4)$.
My attempt: We know that such an $\alpha$ must "look like" $(1432)$, since $(1432)^3=(1234)$. I think I need to find the elements that are conjugate to $(1432)$, so we seek $\alpha$ such that $$\sigma ^{-1} \alpha \sigma = (1432) \implies \alpha = (1\sigma 4\sigma 3\sigma 2\sigma).$$ This seems to me like it means that all permutations of $(1234)$ are fair game because there exists some $\sigma$ that permutes them into $(1432)$. Is this wishful thinking?
Upon taking a power of a permutation, a $k$-cycle can only arise from an $m$-cycle when $k|m$. Since $m$ is limited to $7$ in $S_7$, the $4$-cycle must have arisen from a $4$-cycle. That leaves $3$ other elements, which must be in cycles whose length divides $3$, since otherwise the cycles would survive the cubing. The only possibilities are one $3$-cycle or three $1$-cycles.
If $\alpha$ is a $4$-cycle with $\alpha^3=(1234)$, then $\alpha=\alpha^9=(\alpha^3)^3=(1234)^3=(1432)$. Thus we have no choice for the $4$-cycle; we can only choose to have the other three elements either in three $1$-cycles or in two different $3$-cycles, for a total of $3$ permutations whose cube is $(1234)$:
$$ (1432),(1432)(567),(1432)(765)\;. $$