TITLE: Does a ring isomorphic to its opposite always admit an involution?
QUESTION [5 upvotes]: Let $R$ be a ring with identity such that $R\cong R^{\mathrm{op}}$, then is it necessary that $R$ always admits an involution? An involution is an anti-homomorphism $l\colon R\to R$ such that $l\circ l=\mathrm{id}_R$.

REPLY [3 votes]: Googling "no involution" yielded this for me as the second result:

Morandi, Patrick J., Bharath Al Sethuraman, and J-P. Tignol. "Division algebras with an anti-automorphism but with no involution." (2005).

This is interesting because it leads to projective geometries "with duality but not polarity."
The first proof they give is not constructive, but they do follow up with construction of such division rings, at about Corollary 3.6.  The algebra is involved enough that I don't think I can even fit a sketch here.