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 $S_5$.
I believe "twisted" is being used in a loose sense here. Take a mathematical object, split it into pieces, augment each piece (in a way that is evocative of "twisting") in a way that varies meaningfully with the choice of piece (e.g. smoothly or homomorphically), then put the pieces back together: now you have a twisted version of your original mathematical object. I think this applies to Fourier analysis and tensor products of vector bundles, for example. You could say the Mobius band is a twisted version of a band by this definition, for another example.
In this context, if $H\le G$ is a subgroup and $\phi:H\to C_G(H)$ a homomorphism, then
$$ \{h\phi(h)\mid h\in H\} $$
is an isomorphic copy of $H$ that has been "twisted" by $\phi$ within $G$.
For another example with finite groups, I explained here how the binary octahedral group $2O$ is isomorphic to a "fake $\mathrm{GL}_2\mathbb{F}_3$" (read: twisted version within $\mathrm{GL}_2\mathbb{F}_{3^2}$). You have to interpret "twisting" slightly looser to understand that example, since $\phi$ is really only a homomorphism to $\mathrm{PGL}_2\mathbb{F}_{3^2}$ there, which is why the result is not actually isomorphic to $\mathrm{GL}_2\mathbb{F}_3$.
Best Answer
Hint: They're in bijection with the elements of the set $\{1,2,3,4,5\}$ they don't move.