Is the endomorphism ring of $\mathbb{R}$ self-opposite? Is the endomorphism ring $End\mathbb{R}$ of the Abelian group $\mathbb{R}$ isomorphic to its opposite ring?
All subrings of a self-opposite ring are self-opposite. By choosing an isomorphism of $\mathbb{R}$ with $\mathbb{R}^n$ and observing that linear endomorphisms are also endomorphisms of an Abelian group, we see that $End\mathbb{R}$ contains a copy of the linear endomorphism ring of $\mathbb{R}^n$ for every $n$. However, all of these subrings are self-opposite (by transposition).
My motivation for this is to help me decide on notation for a problem in geometry, which I will not elaborate upon here.

We know that $\mathbb R$ is a direct sum of subgroups isomorphic to $\mathbb Q$; suppose that $I$ is the indexing set of this direct sum. It follows that the endomorphism ring of $\mathbb R$ is the ring of matrices whose columns and rows are indexed by $I$ and whose columns have finitely many non-zero items.
Then you are asking if the ring of column-finite matrices with entries in $\mathbb Q$ is isomorphic to its opposite ring. It seems it isn't: this has been proved in J. J. Simón, There is no analogue for the transpose map for infinite matrices, Publicacions Matemàtiques, Vol 41 (1997), 653–657.
