No, any aútomorphism of $\Sigma_6$ sends $G_i$ to (a conjugate of) $G_i$.

Another relevant post seems to be mathoverflow.net/questions/50864/…

@MaximeRamzi Another thing: The GAP programs in Chris Wuthrichs link only seem to be able to compute the units of $\mathbf{F}_p[G]$ where $G$ is a finite $p$-group (these are called modular group algebras). The Magma program can handle $\mathbf{Z}[G]$ for "small" $G$ (the program uses input on unit groups of $\mathbf{Z}[\zeta_n]$).

@MaximeRamzi Yes, I found the following on Willem de Graafs webpage degraaf.maths.unitn.it/units.html Unfortunately I havent had time to test it yet.....

The paper ”Computing generators of the unit group of an integral abelian group ring” by Faccin, de Graaf and Plesken gives an algorithm which they implement in Magma.

@DavidESpeyer I think you are right: Group #12 on the Shephard-Todd list is isomorphic to $\text{GL}_2(\mathbf{F}_3)$ which does lifts all the way to the $3$-adics.

Let $p(n,k)$ be the sought probability. For $n\geq 1$ obviously $p(n,0)=0$, $p(n,1)=n!/n^n$ and $p(n,n)=1$. Moreover $p(n,n-1)=1-n^{-(n-1)}$. For $n\leq 5$ the computer says that $p(n,2)=\text{A174586}(n)/\binom{n}{2}^n$. OEIS reference is A174586

@PeterKropholler Thanks, I computed this too quickly.... I also found the result 15 for $A_5$ in Hiss' presentation math.rwth-aachen.de/~Gerhard.Hiss/Presentations/… (page 14). However Hiss claims that for $\text{PSL}_2(7)$ the answer is 23, I get 15 in this case as well...

The answer to Richard Borcherds question above appears to be $n^2-n+1$, see D. Eisenbud, "Commutative Algebra: With a View Toward Algebraic Geometry", Exercise 15.33, p. 371. An elementary argument is given in sites.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/gert.pdf

I'm not sure if there is a closed form answer. One natural step would be to compute $p_{n,4}$ which would involve counting solutions to equations in $\Sigma_n$ like $xyz=xzy$ (which is easy) or $xyz=zxy$ (which looks hard...).

@BrendanMcKay: Note that if $x_1\ldots x_k=1$ then any cyclic rearrangment has product $1$ as well. Hence I think the assympotics for $k$ fixed and $n$ large is actually $\frac{(k-1)!}{n!}$.

The probability is given by $1/n!$ for $k=1$ and $2$ and by $\frac{2\cdot n!-p(n)}{(n!)^2}$ (where $p(n)$ is the partition function) for $k=3$.

Rewriting the definition, one gets $f(k,r) = f(k,1)\cdot \frac{\log\log N_k}{(\log\log N_k)+(\log r)}$ so the result follows directly I think.