It's not too difficult to check that there's a bijection between hooks of length $k$ and rim hooks of length $k$. This means that there can be only one rim hook of length $p$. Have a look at Theorem 2.4.7 in the book "The Representation Theory of the Symmetric Group" by James--Kerber for the version of the Murnaghan--Nakayama rule I stated above.

Indeed. It's a good day when I don't personally break representation theory.

I think Theorem 2.7.27 of James--Kerber is relevant here, also the results around there should help.

@AlexeyStaroletov I think it shouldn't be hard to adapt this result to get the general statement. If $w \in \mathfrak{S}_p$ is a $p$-cycle then $\chi^{\alpha}(w) = \mathrm{Res}_{\mathfrak{S}_p}^{\mathfrak{S}_n}(\chi^{\alpha})(w)$. You then have to bound the multiplicities of irreducible constituents of $\mathrm{Res}_{\mathfrak{S}_p}^{\mathfrak{S}_n}(\chi^{\alpha})$, which are LR-coefficients. The bounds on the values of the irreducible constituents are very restricted by the result mentioned by Geoff.

@GeoffRobinson that statement is only true if $\chi^{\alpha}$ is an irreducible character of $\mathfrak{S}_n$ and $w \in \mathfrak{S}_n$ is an $n$-cycle. Then one has $\chi^{\alpha}(w) = 0$ unless $\alpha = (n-r,1^r)$ in which case $\chi^{\alpha}(w) = (-1)^r$. For arbitrary $p$-cycles the value may be different from $\{0,1,-1\}$.

@LSpice Sadly not, it's still sitting on my hard drive. If you want to take a look then I can send you a copy. Just drop me an email.

I like this idea but I think there are some subtleties. I checked with Singular and adding that single term does not generate the radical. Also, how do you know the image of $X$ in the affine algebra is irreducible? It could happen that a polynomial generator becomes reducible in the quotient. For instance $X \equiv (X+1)(Y+1)$ in $K[X,Y]/\langle XY+Y+1\rangle$.

@David. I don't think it's at all clear that the unipotent variety and nilpotent cone aren't isomorphic in the PGL_2 case in characteristic 2. Certainly they are not isomorphic as schemes because one is reduced and the other isn't. However why are their reduced schemes not isomorphic? In other words, why are the underlying varieties not isomorphic? As you say, the nilpotent cone is affine $2$-space $\mathbb{A}^2$ but you only know the unipotent variety is a smooth $2$-dimensional affine variety. I think it's non-trivial to show it's different from $\mathbb{A}^2$.

The point here is that $x$ is unique up to right multiplication by elements of $T_0$.

Yes, one can already see this from the first part of the answer. I'll update the answer to make it clearer.

Well, you need to understand how automorphisms act on unipotent characters of connected groups. This is discussed in several places but you could look at the first chapter of my thesis, link: tel.archives-ouvertes.fr/tel-00709051/document. I needed to use these types of things there. Check out Proposition 1.51, for instance. You could also look at Example 1.76, this is an example of the behaviour you're looking for.

I think the problem is terminology. What I assume Malle means by unipotent is that it is a sum of unipotent characters of $C_{G^{\star}}^{\circ}(s)$. It's not true that a unipotent character of $C_{G^{\star}}(s)$ necessarily restricts irreducibly to $C_{G^{\star}}^{\circ}(s)$. Is this what you're worried is implied?

@MatthiasKlupsch I think from an English perspective I'd slightly modify your formulation as follows: "If $\chi \in \mathrm{Irr}(B)$ is real valued then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi|_{G_{p'}}$." This, for me, is then quite clear.