This question is inspired by [this question][1] about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; but I do not know.
 
Let $G$ be a discrete group, $G^{op}$ its opposite group (i.e. the one with reversed multiplication). Let $H \subset G$ be a subgroup, such that $H=[H,H]=[G,G]$. Note that $H^{op} \subset G^{op}$ has the same properties. I denote the Quillen Plus-Construction of a space with fundamental group $G$ with respect to $H$ by $X \mapsto X^+_H$

> **Question:** Is there a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$, such that the induced map on $\pi_1$ is induced by identity $id: G \to G^{op}$.

Note that there is clearly a homotopy equivalence between $BG_H^+$ and $B(G^{op})_{H^{op}}^+$ such that the induced map on $\pi_1$ is induced by the inverse $inv: G \to G^{op}$; but that is not the one I am looking for.

However, this shows that one could also ask:

> **Question:** Is there a homotopy equivalence between $BG_H^+$ and itself, such that the induced map on the abelian group $\pi_1(BG_H^+) = G/H$ is the inversion.

Given the motivation, any good answer in the case $G=GL_{\infty}(R)$ (for some ring) would be interesting too.

  [1]: http://mathoverflow.net/questions/46597/why-does-the-grothendieck-group-k-0r-of-a-ring-not-depend-on-our-choice-of-us