I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
This is more delicate and difficult than most problems about Sylow subgroups, and it must be tackled by relating the 5-sylow subgroups and the 3-sylow subgroups. In general such a problem in group theory at the undergraduate level is either going to be trivially easy, or require this technique.
As you established $n_3 = 1$ is one possibility but it was noted in the comments that $n_3 = 10$ is also possible. Suppose $n_5 = 6$ so that the $5$-sylows are not normal. Then they must be their own normalizers because the number of them is equal to the index of the normalizer of any one, and they are the unique subgroups of order $5$. However the fact that only $5$ divides $30$ and not $25$ allows us to know that all of the elements of order $5$ in this group are contained in a $5$-sylow, and that these sylows have trivial intersection. But that means there are $(5-1)*6 = 24$ elements of order $5$ in this group. But this means that there cannot possibly be $10$ subgroups of order $3$ otherwise there would be $20$ elements of order $3$ and that would give $47$ elements in the group which would be absurd.
So we know that the $3$-sylows would have to be normal in this situation. Then we have that if $H_3$ is a $3$-sylow and $H_5$ is a $5$-sylow then $H_5H_3$ is a subgroup because $H_3$ is normal. But then $H_5H_3/H_3 \cong H_5/(H_5 \cap H_3)$ and counting orders we know that $H_5H_3$ has order $15$. But there is a unique group of order $15$ (there are many ways to show this, but in our particular situation the fact that $H_5H_3 \cong \mathbb{Z}_{15}$ comes down to the fact that $H_5$ is normal by a sylow argument on this subgroup) and thus this group is abelian and this contradicts that $H_5$ is its own normalizer. Thus $H_5$ is normal.
Finally we can now count again. Since $H_5$ is normal we can take any one of these $H_3$ and do the same argument as the above to realize that $H_3$ has a normalizer of order at least $15$ and thus $H_3$ must be normal because it is supposed to have normalizer or index either $1$ or $10$.
Hope this helps.
Best Answer
There are indeed $\frac156!=144$ different $5$-cycles in $S_6$ but note that a single $5$-Sylow accounts for 4 of them.
Indeed $c$, $c^2$, $c^3$, $c^4$ are the 4 different non-trivial elements in $\langle c\rangle$.
Thus the total number of $5$-Sylows is $\frac14144=36$.