Is the group von Neumann algebra construction functorial?

Well, for $sin G$ let $lambda(s)$ be the left-translation operator by $s$; all such operators are in the group von Neumann algebra. I guess that the hoped for homomorphism $F:NG rightarrow NH$ should satisfy $F(lambda(s)) = lambda(f(s))$ for $sin G$, and we should have that $F$ is an (ultraweakly) continuous $*$-homomorphism. In particular, $F$ is contractive.

Then $F$ need not exist. Let $G=mathbb Z$ and $H=mathbb Z/nmathbb Z$. Then $NH = CH$, so we have a trace on $NH$ which sends $lambda(0)$ to $1$. So if we apply $F$, and then take the trace, we should get an ultraweakly continuous functional $phi$ on $NG$ which satisfies $phi(lambda(ns)) = 1$ for all $sinmathbb Z$.

But this can't happen: maybe we can see this via the Fourier transform. Then $NG cong L^infty(mathbb T)$ and $phi$ induces $hin L^1(mathbb T)$ which satisfies $int h(theta) e^{instheta} dtheta = 1$ for all $sinmathbb Z$. This violated Reimann-Lebesgue.

On the other hand, if $G subseteq H$ is an inclusion (of discrete groups, to avoid topology) then we do get an inclusion $NG rightarrow NH$. Here's a construction. Find an index set $I$ and $(h_i)$ in $H$ such that $H$ is the disjoint union of ${Gh_i}$. Then define $V:l^2(H)rightarrow l^2(G)otimes l^2(I)$ by $V(delta_h) = delta_gotimesdelta_i$ if $h=gh_i$ (so defined on point-masses, and extend by linearity). So $V$ is unitary, and $theta:xmapsto V^*(xotimes 1)V$ is a normal $*$-homomorphism $NGrightarrow B(l^2(H))$. Then, for $rin G$, $V^*(lambda(r)otimes 1)V(delta_h) = V^*(delta_{rg}otimesdelta_i) = delta_{rh}$ as $rgin G$. So $theta$ maps into $NH$, and does what we want.

Surely there is some general result, but I'm not sure of it...