How many different elements can we obtain by multiplying all element in a group?

Let $G$ be a finite group.

How many different elements can we obtain by multiplying all element in a group?

Of course, if $G$ is abelian the answer is one but when G is non-abelian, changing the order of the multiplication may produce new elements.

My second question is actually related to my attempt to solve the first one.

Let $S$ be set of all elements produced by multiplying all elements in $G$. Then, it is easy to show that $Aut(G)$ acts on $S$ naturally. I wonder whether this can be transitive.

All products are equal modulo the commutator subgroup, so $S$ is contained in a coset of $G'$. It turns out that $S$ is equal to this coset:

http://www.sciencedirect.com/science/article/pii/S0304020808732572

So the answer is $|G'|$.

The answer to your question is even more subtle. The set of all the possible products is always a coset of the commutator subgroup.

Theorem Let $G$ be a finite group of order $n$, say $G=\{g_1, \dots, g_n\}$ and let $P(G)=\{g_{\sigma(1)}\cdot g_{\sigma(2)} \dots g_{\sigma({n-1})} \cdot g_{\sigma(n)}: \sigma \in S_n\}$.

(a) If $|G|$ is odd, then $P(G)=G'$

(b) If $|G|$ is even, let $S \in Syl_2(G)$. Then $P(G)=G'$ in case $S$ is non-cyclic. If $S$ is cyclic, then $P(G)=xG'$, where $x$ is the unique element of order $2$ of $S$.

This higly non-trivial and beautiful result relates to combinatorics - the construction of Latin Squares. It heavily relies on the proof of the so-called Hall-Paige conjecture (Hall, Marshall; Paige, L. J. Complete mappings of finite groups. Pacific Journal of Mathematics 5 (1955), no. 4, 541--549.). This could be proved thanks to the classification of the finite simple groups.