When the elements of maximum order are $n$-cycles in $S_n$? If the elements of maximum order in $S_n$ are $n$-cycles, then we can guess with few computations that $n$ must be at most $4$. How can we prove this?
I tried the case in $S_{2n+1}$, the symmetric group on odd number of letters. Here, the element $(1,2,...,n)(n+1,n+2,\cdots, 2n+1)$ is product of two disjoint cycles of relatively prime order- $n$ and $n+1$, and so its order is $n(n+1)$. Then $n(n+1)\leq 2n+1$ implies that $n\leq 1$, i.e. the symmetric group under consideration will be $S_1$ or $S_3$.
How to handle the case of symmetric group $S_{2n}$ on even number of letters?
 A: In the case of $S_{2n}$, your argument almost goes through in most cases! Take $S_{2n-1} \subset S_{2n}$ (considered as the subgroup of permutations fixing $n$) and use your construction to obtain an element of order $(n)(n-1)$. In this case, $n(n-1) \leq 2n$ implies that $n^2 \leq 3n$, which fails for $n \geq 4$.
This leaves just the case of $S_6$. By general results on symmetric groups, any cycle can be expressed as a product of disjoint cycles. The order is determined by the cycle type, which in turn is determined by a partition of $6$ (the partition corresponds to the lengths of the different cycles appearing in the cycle decomposition of the element). The possible partitions are:
$$(1,1,1,1,1,1)$$
$$(2,1,1,1,1)$$
$$(2,2,1,1)$$
$$(2,2,2)$$
$$(3,1,1,1)$$
$$(3,2,1)$$
$$(3,3)$$
$$(4,1,1)$$
$$(4,2)$$
$$(5,1)$$
$$(6)$$
In each case, the order of an element of cycle type given by the partition is the LCM of the numbers in the partition. Checking all cases, we see that this never exceeds $6$ and indeed $6$-cycles in $S_6$ are of maximal order. However, elements of cycle type $(2,3,1)$ are also of maximal order.
A similar (but easier) partition check shows that if $n \in \{1,2,3,4\}$, then $n$-cycles are of maximal order (equal to $n$), and no other elements are of order $n$.
