This question is inspired by this question 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.

EDIT: Johannes Ebert suggested in a comment a strategy to give a negative answer to Question 2 in general. The Kan-Thurston construction gives a way of obtaining every (finite?) cell complex $X$ as the plus-construction on $BG$ with respect to some perfect subgroup $H$. It seems that the only missing piece is a finite cell-complex $X$ with $\pi_1$ abelian and no $\pi_1$-inversion-inducing self-homotopy equivalence. Then, $X = BG^+_H$ for some group $G$ and some perfect subgroup $H \subset G$. However, since $\pi_1 = G/H$ is abelian, we see that $[G,G] \subset H$ and hence, $[H,H] = [G,G]=H$ as required.

Question: Can anybody give an example of such a space?