Given a polynomial parametrized by $p(t)=(x(t),y(t))$ such that $y(t)=p(t)$, $x(t)=t$, and $p(t)= \sum_{i=0}^na_it^i$, for real coefficients $a_i$, is there some transformation I can take such that the curve becomes "straightened out" into a flat horizontal line. So, I'm looking for an transformation that turns the slope of the polynomial to zero and preserves arclength. I know Möbius transformations are distance preserving maps, so maybe an inverse Möbius transform would do? A log transform won't do because that doesn't set the slope to zero, although it does "straighten out" monomials, and that isn't arclength preserving anyway. I looked up isometry groups, but all I really found was information on symmetries, rotations, and translations. I suspect I need some complex or hyperbolic isometry, but I'm not sure since I'm using a formula for a polynomial in Euclidean space.
So, for example, such a transformation would take $p(x)=x^2$ and map $p(0)=0$ to $0$, and since the length of $p(x)$ along $[-1,1]$ is about $2.96$, it would set $p(x)$ equal to the $0$ map along the interval $[-1.48,1.48]$.
There is such a transform, but I expect you're not going to like it. Define
$$ G(x, y) = (x, y - p(x)) $$ When applied to the point $(x, p(x))$, this produces $(x, 0)$, so the graph of $f$ gets straightened out into the $x$-axis by $G$.
You might think "No, I meant some transform in a limited class of geometric transforms, like rotations, or mobius transforms, or nonsingular linear transformations," but all of those have the property that they preserve lines (either in the plane or in projective space). But the graph of a nonconstant degree-$n$ polynomial has the property that there may be a line $L$ that intersects it in $n$ distinct points. If you could transform this graph to a straight line, then you'd have to transform $L$ to something that intersects that straight line $n$ times...which can't possibly be a line. So all line-preserving transforms are hopeless.
If you want to clarify what you consider a "transform," I'm happy to enlarge on this answer a bit.
Addition The revised question says that you're looking for an isometry. That's completely hopeless.
Consider, for instance, $y = x^2$, and look at the points $A = (-1, 1)$, $ B = (0, 0)$, and $C = (1, 1)$. If you've got an isometry $F$ of $\mathbb R^2$ that carries the graph of $y = x^2$ to a straight line, then you can assume that the line is the $x$-axis, and that $B$ is left fixed (by composing with a translation and rotation of the plane).
Now the point $F(A)$ lies on the real axis, and its distance from $B$ must be the same as the arclength from $A$ to $B$ along the original graph (because every isometry is a local isometry, so arclengths are preserved). That arclength, by basic calculus, is greater than $1$ (indeed, greater than $\sqrt{2}$). The same goes for $F(C)$; and these two must have $F(B)$ between them. So the distance from $F(A)$ to $F(C)$ is greater than 2. On the other hand, the distance from $A$ to $C$ in the domain of $F$ is exactly $2$. Hence $F$ cannot be an isometry.
Since you can't find an isometry to straighten even this one quadratic, it's hopeless to look for isometries to straighten out anything more complicated.
Post-comment addition
The OP has clarified that what's wanted (I believe) is a transform $G$ with the property that corresponding arcs of the graph of $p$ and the graph of $G \circ f$ should have the same length, i.e., a homeomorphism (or difeomorphism) of $\mathbb R^2$ that's arclength preserving on the graph of $p$.
If we let
$$ A(t) = \int_0^t \sqrt{1 + p'(t)^2} ~ dt $$ then we can define such a map by saying $$ G(x, y) = (A(x), y - p(x)) $$ This map, like the one above, takes points with $ y = p(x)$ to points on the $x$-axis, but ensures that their position on the $x$-axis is indeed their arclength from $(0, p(0))$, so that the map is arclength preserving.
For all but the simplest polynomials $p$, the function $A$, although well-defined and differentiable by the fundamental theorem of calculus, will be impossible to express in "elementary terms" -- you won't be able to compute the antiderivative with any of the tricks you learn in you calculus book.
On the other hand, if you're interested in using this to do something cute in a video game, numerical integration, tabulated sufficiently finely, might work to get a picture that's visually indistinguishable from the correct picture.