Say, we have a countable ICC group $G$, a Hilbert space $H$ with a basis indexed by the group elements, the group algebra generated by the left regular representation of $G$ on this Hilbert space, and its norm and weak closures, the reduced C* and the von Neumann algebras.
On the other hand, an automorphism of $G$ also permutes the basis of $H$ (when applied to the indices) and thus defines a unitary on $H$.
What's the relationship of this unitary to the group algebras? Is it ever in, say, the von Neumann group factor? How do the properties of this unitary reflect whether it comes from an inner group automorphism or not?
These unitaries, representing automorphisms of $G$, also give rise to an operator algebra. How is this algebra related to the group's, say, von Neumann algebra?

1$\begingroup$ This is a collection of interesting questions, IMHO. But the questioner could have done some basic work before asking. For example, the group von Neumann algebra is the commutant of the right regular representation, and this gives a criteria for these "automorphism unitaries" to be members of the group von neumann algebra. As far as I can see, this is only the case for the identity automorphism. $\endgroup$ – Matthew Daws Jun 20 at 9:30
If I interpreted your question correctly, what you are interested in is the following result:
Suppose $G$ is an i.c.c., discrete group and $\delta: G \rightarrow G$ is a group homomorphism. Define the automorphism of the group von Neumann algebra $\theta(\sum_g c_g \lambda_g)= \sum_g c_g \lambda_{\delta(g)}$. Then, $\theta$ is outer if and only if $\delta$ is an outer automorphism of $G$.
A reference for the aforementioned result is Remark 2.3 in the paper " A generalization of free action" by Robert R. Kallman.

$\begingroup$ I think the OP was rather asking to define a unitary $U$ by $U(e_g) = e_{\delta(g)}$ where $(e_g)$ is the standard o.n. basis of $\ell^2(G)$. Then the questions is about properties of $U$ and $\delta$. $\endgroup$ – Matthew Daws Jun 20 at 20:33

$\begingroup$ Thanks, Darth, yes I am. However, in my question the setup is different. In the result you mentioned, the group automorphism acts on the group, which then maps to its representation, which then completes to a von Neumann algebra. So one direction is obvious  if the group automorphism is a conjugation by a group element, then conjugation by that element's corresponding unitary in the left representation will be our automorphism on the algebra level. In my question, the automorphism permutes the Hilbert space basis and becomes another unitary, likely outside of the algebra. $\endgroup$ – Chilperic Jun 20 at 20:57

$\begingroup$ @MatthewDaws: wouldn't ad(U) just implement the automorphism $\theta$ that I defined in my answer? $U$ is in $L(G)$ if and only if $\theta$ is inner, in that case, right? Am I missing something? $\endgroup$ – Darth Vader Jun 20 at 20:58

$\begingroup$ @Chilperic: Yes one direction iss obvious. The other direction is that $\delta$ is outer implies $\theta$ is outer. This is what's proved in Remarks 2.3 in the quoted paper. The proof is basically what Matthew pointed out in the comment to your question. $\endgroup$ – Darth Vader Jun 20 at 21:00

$\begingroup$ Sorry, Darth, posted before edits finished. Basically I m curious to what extent $Aut(G)$ and $Out(G)$ properties extend to the von Neumann factor. Can we even tell from an automorphism of the factor whether it comes from a group automorphism or not? $\endgroup$ – Chilperic Jun 20 at 21:05