This is from a practice exam (paraphrasing):
Let $\chi$ be a character of $S_6$, given by $\chi(g) = |\text{Fix}(g)|-1$ for the action of $S_6$ on the set of its sylow $5$-subgroups. Decompose the restriction of $\chi$ to $S_5$ into irreducibles.
Remark: I presume that action is by conjugation.
Here is my reasoning: The sylow $5$-subgroups has to be cyclic groups of order $5$ generated by a $5$-cycle. There are $\binom{6}{5} 4! = 6 \cdot 4 \cdot 3 \cdot 2$ $5$-cycles, and each $5$-sylow subgroup has $4$ non-identity elements, so I believe we can divide by $4$. Hence I claim there are $\frac{6 \cdot 4!}{4} = 6 \cdot 3 \cdot 2 = 36$ sylow $5$-subgroups.
We just have to think what type of permutations there are in $S_6$. We have the following types (using representative elements): $(1),(12),(123),(1234),(12345),(123456),(12)(34)(56),(12)(34),(123)(45),(123)(456),(1234)(56)$.
From what we said above, for arbitrary $P \in \text{Syl}_{5}(S_6)$ we have $P = \langle \sigma \rangle$ for some $5$-cycle $\sigma = (abcde)$.
So then $$P = \{1,(abcde),(acebd),(adbec),(aedbc)\}.$$ For $\tau \in S_6$ to fix this, under action by conjugation, means that $\tau \in N_{S_6}(P)$.
Since we make the restriction to $S_5$, it is enough to know how elements in $S_5$ acts on the set $\text{Syl}_{5}(S_6)$. Since a character is constant on conjugacy classes, it is enough to take a representative element from each conjugacy-class in $S_5$.
In $S_5$ we have the conjugacy-classes $(1),(12),(123),(1234),(12345),(12)(34),(123)(45)$. From now on, let $\psi$ be the restriction of $\chi$ to $S_5$. Clearly $(1)$ fixes all elements in $\text{Syl}_{5}(S_5)$, so $\psi(1) = 36-1 = 35$.
Note that by orbit-stabilizer, we have $36 = \frac{6!}{|N_{S_6}(P)|} \iff |N_{S_6}(P)| = \frac{6!}{6^2} = 20$. From Let $P \in \text{Syl}_{5}(S_5)$. Show that the normalizer $N := N_{S_5}(P)$ is a monomial group. we see that $N_{S_6}(P) = HP$ where $H = \langle \tau \rangle$ for $\tau$ a $4$-cycle, since the same reasoning applies here as there (I believe). We see from this that no $3$-cycle normalizes any $P$, so $\psi((123)) = -1$.
What about $2$-cycles? Well, there are no $2$-cycles in any $N_{S_6}(P)$ for any $P$. This follows from reasoning in the link in the previous paragraph; recall that permutations in $S_n$ are only conjugate iff they are of the same cycle-type, and the only if direction is preserved in subgroups. One finds that no conjugacy-class is represented by a $2$-cycle, in $N_{S_6}(P)$. So $\psi((12)) = -1$.
What about $(1234)$. Well, clearly $\tau_{P}$ associated with $N_{S_6}(P)$ normalizes $P$. Can $\tau_{P}$ normalize any other sylow $5$-subgroup $P'$? I believe the answer is conclusively no, since $\tau$ is completely determined by the generator $\sigma$ for $P$, which trivially are different for different sylow $5$-subgroups. So I claim that $\psi(\tau_P) = 0$.
By Steve D:s comment, I realize the reasoning here is wrong. We have $\binom{6}{4}3! = 90$ $4$-cycles, and one finds that each normalizer of some $P \in \text{Syl}_{S_6}(P)$ has, I believe, $10$ $4$-cycles. So we need $360$ cycles. So, by hypothesis, we should have that each $4$-cycle normalizes $\frac{360}{90} = 4$ $5$-sylow subgroups. The issue here I see is the same as for $(12)(34)$, I don't know how to argue that they distribute equally over all the normalizers.
For $5$-cycles, say $(12345)$, then clearly $(12345)$ normalizes $P = \langle (12345) \rangle$. Again, from the link earlier, we see that no other element in $N_{S_6}(P)$ is a $5$-cycle, except $\sigma$ and it's powers. So I claim that $\psi((12345)) = 0$.
For $(123)(3)$, we see that the order is $6$. But no element of order $6$ can be in $N_{S_6}(P)$ since $6 \nmid |N_{S_6}|(P)| = 20$. So $\psi((123)(45)) = -1$.
Finally, for the conjugacy class represented by $(12)(34)$. In $S_6$, there are $\frac{\binom{6}{4}\binom{4}{2}}{2!} = 45$ such elements, and we have $36$ sylow $5$-subgroups, with associated normalizer $N_P$, each containing $5$ elements of cycle type $(2,2)$ (see the link). So we need $5 \cdot 36 = 180$ such "elements". Notice that $\frac{180}{4} = 45$. So we hypothesize that each permutation of cycle type $(2,2)$ in $S_6$ normalizes $4$ sylow $5$-subgroups.
Assume that $\theta = (25)(34)$ normalizes $P_1,\ldots,P_r$. Let $y = x\theta x^{-1}$. Then $y(xP_ix^{-1})y^{-1} = (x\theta x^{-1})(xP_ix^{-1})(x\theta^{-1}x^{-1}) = x\theta P_i\theta^{-1}x^{-1} = xP_ix^{-1}$. So $y \in N_{S_6}(xP_ix^{-1})$. Notice that $x$ was arbitrary. I am not entirely sure what to conclude from this last part though. I'll post this as a partial answer (due to constrained time).
I'd be happy to receive any feedback on my partial answer, corrections and also a way to show that elements in $S_5$ with cycle-type $(2,2)$ normalizes $4$ sylow $5$-subgroups, so fixes $4$ elements under the action, i.e. $\psi((12)(34)) = 3$.
Update: So, again, by Steve D:s comment, which I believe is basically the same point I made in the paragraph regarding $(25)(34)$:
Let's focus on the $4$-cycles. I claim that exactly the same argument will work for the case $(12)(34)$. As we've seen, if $\theta$ normalizes the sylow $5$-subgroups $P_1,\ldots,P_n$, then $x\theta x^{-1}$ normalizes $xP_1x^{-1},\ldots,xP_nx^{-1}$. Note that $x\theta x^{-1}$ has to be an element of the same cycle type as $\theta$, and that all elements of the same cycle type are conjugate. Since $x$ was arbitrary, we can get any element of the same cycle type as $\theta$ by conjugating with suitable element $x \in S_6$. So this means that if some $\theta$ normalized $\geq 5$ sylow subgroups, then all elements of the same cycle type would normalize $5$ sylow $5$-subgroups. In the case of $4$-cycles, we saw that there was $90$ $4$-cycles, and each $P$ had $10$ $4$-cycles. But now we have $5 \cdot 90 = 450$ $4$-cycles to distribute on $36$ $P$ groups. This will lead to a contradiction, since we are only allowed to have $10$ $4$-cycles in one normalizer. If there was some $\theta$ that normalized less than $4$ sylow $5$-subgroups $P$, then all elements of the same cycle type would have to normalize $<4$ (and then we don't get $360$ elements of the same cycle type as $\theta$ in the normalizers); if there was one $\xi$ of the same cycle type as $\theta$ that normalized $4$ or more, then every other element would have to do that, by what we just said, so in particular $\theta$ (contradiction!). This argument works for both $(12)(34)$ and $(1234)$, I $\underline{\text{claim}}$.
