A group $G$ is said to have the Bergman property (the property of uniformity of finite width) if given any generating $X$ with $X=X^{-1}$ of $G$, we have that $G=X^k$ for some natural $k$, that is, every element of $G$ is a product of at most $k$ elements of $X$. We prove that the automorphism group $\operatorname{Aut}(N)$ of any infinitely generated free nilpotent group $N$ has the Bergman property. Also, we obtain a partial answer to a question posed by Bergman by establishing that the automorphism group of a free group of countably infinite rank is a group of uniformly finite width.

The automorphism group of a finitely generated free group is the normal closure of a single element of order 2. If $m<n$, then a homomorphism ${\rm Aut}(F_n)\to {\rm Aut}(F_m)$ can have image of cardinality at most 2. More generally, this is true of homomorphisms from ${\rm Aut}(F_n)$ to any group that does not contain an isomorphic image of the symmetric group $S_{n+1}$. Strong restrictions are also obtained on maps to groups that do not contain a copy of $W_n=({\mathbb Z}/2)^n\rtimes S_{n}$, or of ${\mathbb Z}^{n-1}$. These results place constraints on how ${\rm Aut}(F_n)$ can act. For example, if $n\ge 3$, any action of ${\rm Aut}(F_n)$ on the circle (by homeomorphisms) factors through ${\rm det} : {\rm Aut}(F_n)\to{\mathbb Z}_2$.The research of the first author was supported by an EPSRC Advanced Fellowship. The research of the second author is supported in part by NSF grant DMS-9307313.