What surfaces in $\mathbb R^3$ are such that every planar section (with more than 1 point) has nontrivial symmetry? In $\mathbb R^3$ , the intersection of a plane and a sphere (e.g. $x^2 + y^2 + z^2 = 1$) is either empty, a single point, or a circle.  All isometries of those circles are realized by isometries of the full sphere. In contrast, every plane intersects a cone (e.g. $x^2 + y^2 - z^2 = 0$) in a conic section which has reflection symmetries along one, two, or more axes.  The one reflection is realized by an isometry of the cone, but the second, in general, is not. 
What surfaces in $\mathbb R^3$ , or general subsets of $\mathbb R^3$ , are such that all non-empty planar intersections are either 1 point or have nontrivial symmetry?  When are those symmetries not extensible to a symmetry of the full surface?
 A: I suspect the answer is that the surface in $\mathbb R^3$ must be a homogeneous quadratic form in $(x,y,z)$ set equal to a constant, as in
$$ a x^2 + b y^2 + c z^2 + r y z + s z x + t x y = k.     $$
For you, all of $a,b,c,r,s,t,k$ are real numbers.  There is little benefit to allowing lower order terms as in $\alpha x + \beta y + \gamma z,$ a translation takes the result to center on the origin again. For that matter, if you know enough linear algebra, a rotation takes this surface into a diagonalized one,
$$  a_1 x_1^2 + b_1 y_1^2 + c_1 z_1^2 = k_1,    $$ where some of $a_1,b_1,c_1,k_1$ may be positive, some $0,$ some negative. You might try graphing these with all $a_1,b_1,c_1,k_1 \in \{-1,0,1  \},$ there are 16 possibilities but there is repetition, so maybe take $a_1 \leq b_1 \leq c_1$ but any $k_1.$
Note that the intersection of any plane with this surface is, in coordinates appropriate for that plane, and orthogonal coordinates if we demand it, a quadratic in two variables $u,v$ call it $A u^2 + B u v + C v^2 + D u + Ev = F.$ This has symmetries, or may be a single point, or a line or pair of lines, and so on. 
Well, these examples work. I do not expect there will be any others. For most planes through one of these surfaces, the symmetry of the figure within the plane will not extend to the whole surface. Finally, a proof that only these surfaces work would be pretty elaborate.  
EDIT: user mjqxxxx gave some I had not considered, the one that is connected and new is: take any curve in the $xy$ plane that has a symmetry, a reflection or $180^\circ$ rotation or something. Then construct the cylinder over it, meaning take the same figure with arbitrary $z.$  Any plane slice preserves the symmetry. Now, my current opinion is that if we take something in the plane that has only a $120^\circ$ rotation symmetry, some kind of pinwheel, after making the cylinder most slanted planes through the cylinder will not preserve that symmetry. Need to thinnmmnnkk. That would be a nice result, though, connected examples are either full-dimensional quadratic things or cylinders over lower-dimensional examples. EDDITTT: more in comments below. This is getting out of hand. The second example that is essentially lower-dimensional is take any curve in the $xy$ plane with $y \geq 0$ and rotate it around the $x$-axis. 
