Find all the isometries which fix a conic Given a conic on $\mathbb{R^2}$, for example $\mathscr{E}: 4x^2+y^2=4$, just to have a simple one, how to find all the isometries of $\mathbb{R^2}$ which fix $\mathscr{E}$? And all the affinities?
To find an isometry or an affinity is quite simple, but all of them gives me a hard time.
Thank you!
 A: The isometries that fix a conic are its symmetries. For a central conic, this group is generated by the reflections in its principal axes; there are four such isometries unless you’re working with a circle, in which case all isometries that fix the circle’s center leave the circle fixed. For a parabola, there are only two: the identity map and the reflection in its axis.  
Opening the problem up to any affine transformation yields a richer set. Since every ellipse is related by an affinity, ditto hyperbolas and parabolas, w.l.o.g. we need examine only the unit circle, the “unit” hyperbola $x^2-y^2=1$ and the parabola $y^2=x$.  
Circle: The center is a fixed point of the transformation, so only linear transformations need apply. Will Jagy explores this case in this answer: the linear transformations which fix the unit circle are exactly the isometries that leave the origin fixed, i.e., all orthogonal transformations. The conjugates of these by an affine transformation that maps the unit circle to a given ellipse are then the affine transformations that fix that ellipse.  
Hyperbola: Similarly to the case of the circle, we’re looking for linear transformations that preserve the quadratic form $x^2-y^2$. These turn out to be Lorentz transformations, and as with the previous case, we can generate the transformations that fix an arbitrary hyperbola by conjugating them with the appropriate affine transformation.  
Parabola: I don’t know of a distinguished group that fixes the parabola $y^2=x$ and a quick search didn’t turn anything up, but it’s not too hard to come up with a set of affine transformations that meet this criterion. The matrix of this conic (using the convention that the line at infinity has homogeneous coordinates $[0:0:1]$) is $$C=\begin{bmatrix}0&0&-\frac12\\0&1&0\\-\frac12&0&0\end{bmatrix}.$$ After transformation by a generic affine transformation, the resulting matrix must be a nonzero scalar multiple of $C$. If you work through the resulting constraints, you will end up with the two-parameter family of transformations with matrices of the form $$A = \begin{bmatrix}a^2 & 2at & t^2 \\ 0 & a & t \\ 0&0&1 \end{bmatrix}.$$ Examining the eigenvectors of $A$ is instructive. One is obviously $[1:0:0]$—the intersection of the parabola’s axis with the line at infinity (effectively the axis direction vector). This says that the axis direction of the transformed parabola is the same as the original, as we’d like. The other two eigenvectors are $[t^2:(1-a)t:(1-a)^2]$ and $[2t:1-a:0]$. The first of these turns out to be a point on the parabola, so the transformation fixes a particular point on $\mathscr C$. The second is the intersection of the tangent at this fixed point with the line at infinity—the direction of this tangent, essentially—which is therefore also preserved by the transformation. Two distinct points and the tangents at those points are enough to determine a parabola uniquely, so these invariants do indeed correspond to the same parabola, $y^2=x$. For an arbitrary parabola, we can either conjugate $A$ with the appropriate affine transformation, or construct one directly from the parabola’s axis direction, a point on the parabola and the direction of the tangent at that point.

Update: The invariants of the parabola-preserving affine transformation can be stated in a slightly different way. It has a fixed point on the parabola and three invariant lines: the line at infinity (as do all affine transformations), the tangent at the fixed point, and the line through the fixed point parallel to the parabola’s axis. The matrix $A$ can be derived from these invariants. We can also come up with a more geometrically meaningful parameterization of $A$. If the fixed point is $[y^2:y:1]$, we can write $$A = \begin{bmatrix}a^2 & 2ya(1-a) & y^2(1-a)^2 \\ 0 & a & y(1-a) \\ 0 & 0 & 1\end{bmatrix}.$$ The eigenvectors of this matrix are $[y^2:y:1]$, $[2y:1:0]$ and $[1:0:0]$ with respective eigenvalues $1$, $a$ and $a^2$. It’s easy to see from this eigensystem that $A$ represents scaling by a factor of $a^2$ in the axis direction and by $a$ in the direction of the tangent at $[y^2:y:1]$, followed by a translation to compensate for any displacement of that point.
