If I consider an elliptic curve $C$ as a Riemann surface cut out in $\mathbb{C}P^2$ by a homogenous cubic, and if that cubic is defined over $\mathbb{R}$, then I think we have a conjugation map $C\rightarrow C$ that is an isomorphism if we regard $C$ as a 2-dimensional real manifold (where we just take the coordinates and take their complex conjugates).
Also, the fixed points of that isomorphism should be the points of $C$ that are also in $\mathbb{R}P^2$.
Unlike the complex picture, it is easy to graph the real points of $C$ (using http://www.math.uri.edu/~bkaskosz/flashmo/implicit/ for example). If our cubic is $x^3-x=y^2$ (in an affine piece), then I got something that looks topologically like 2 circles. If our cubic is $x^3-x+1=y^2$, then I got something that looks topologically like one circle.
I can believe there is an involution of the torus where the fixed points is 2 circles (just reflect it), but the second example suggests there is some involution where the fixed points is 1 circle. Do you know what that involution should look like?
Edit: This question was motivated by the diagram under example M of chapter 2 of Mumford's Red Book of Varieties and Schemes.
If $\alpha \in \mathbb{C}$ is such that $\operatorname{Re}(\alpha) > 0$ and $\operatorname{Im}(\alpha) > 0$ then complex conjugation has a single circle as fixed point set on the elliptic curve $\mathbb{C}/(\mathbb{Z}\alpha + \mathbb{Z}\overline{\alpha})$.