Is there a characterization of automorphisms of degree-$2$ affine plane curves over $\mathbb{R}$? Let $A$ denote a (nonsingular, irreducible) affine plane curve over $\mathbb{R}$ given by the explicit equation
$$A : a_Ax^2+b_Axy+c_Ay^2+d_Ax+e_Ay+f_A$$

Question. Is there an explicit description of the automorphisms of $A$ in the category of real affine varieties?

 A: We will compute the result for $\Bbb C$, and note that an automorphism of $\operatorname{Spec} \Bbb R[x,y]/f$ is the same as an automorphism of $\operatorname{Spec} \Bbb C[x,y]/f$ with real coefficients. Over $\Bbb C$, there are exactly two affine plane conics up to isomorphism: $xy=1, y=x^2$. So it suffices to compute the automorphism groups of each of these curves and do some work translating things back to the reals.
One preliminary result that I'll use without proof: $Aut(k(x))= PSL(2,k)$ acting by Mobius transformations.
Case 1: $y=x^2$. The coordinate algebra is isomorphic to $\Bbb C[x]$ by writing $y=x^2$. Any automorphism of $k[x]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x]$ are exactly the linear maps. So the automorphisms of $\Bbb C[x]$ are exactly the linear maps $x\mapsto ax+b$ with $a\neq 0$. In order for these automorphisms to restrict the correct way to our curve, we just need $a,b\in\Bbb R$.
Case 2: $xy=1$. The coordinate algebra is isomorphic to $\Bbb C[x,x^{-1}]$, and any automorphism is determined by it's action on $x$. Again, any automorphism of $k[x,x^{-1}]$ extends to an automorphism of $k(x)$, and the automorphisms of $k(x)$ that preserve $k[x,x^{-1}]$ are $x\mapsto ax$ and $x\mapsto cx^{-1}$ for $a,c\neq 0$. 
We need to do a little more work in case 2 to translate these automorphisms back to the land of the reals. In the case that our curve over $\Bbb R$ is a hyperbola, we can make the change of variables $x\mapsto (x-y),y\mapsto (x+y)$ to translate between $xy=1$ and $x^2-y^2=1$. Recognizing this coordinate change as a rotation and performing some algebra, we see that the symmetries of $x^2-y^2=1$ are all of the form $\begin{pmatrix} \pm \cosh(a) & \sinh(a) \\ \pm \sinh(a) & \cosh(a) \end{pmatrix}$ for $a\in [0,2\pi)$.
Similar logic with the change of variables $x\mapsto x-iy, y\mapsto x+iy$ shows that the automorphisms of $x^2+y^2=1$ are exactly the rotation matrices up to a reflection across any line through the origin, ie matrices of the form $\begin{pmatrix} \pm\cos(a) & -\sin(a) \\ \pm\sin(a) & \cos(a) \end{pmatrix}$.
I wouldn't be surprised if there were a shorter proof, but this is the clearest approach I could think of while writing. 
