Doesn't Newton's equation of motion have a bigger invariance group than the Galilean group? Newton's equation ${F}^i=m\frac{d^2x^i}{dt^2}$ is unchanged in form, under the Galilean group:

*

*(i) under a translation of the origin of coordinates,


*(ii) rotation of coordinates, and


*(iii) Galilean boosts.
If we however consider the general coordinate transformation of the form $x'^i=x'^i(x^1,x^2,x^3)$, for example, going from Cartesian to spherical or cylindrical coordinates,  doesn't the form of Newton's also remain unchanged in the sense that if $F^i=m\frac{d^2x^i}{dt^2}$ holds in the Cartesian coordinates, ${F^i}^\prime=m\frac{d^2x^{\prime i}}{dt^2}$ holds in the Spherical or cylindrical coordinates, where $F^i$ and $\frac{d^2x^i}{dt^2}$ transform as $$F^{\prime i}=\frac{\partial x^{\prime i}}{\partial x^j}F^j, \quad \frac{d^2x^{\prime i}}{dt^2}=\frac{\partial x^{\prime i}}{\partial x^j}\frac{d^2x^j}{dt^2}.$$
So am I justified in saying that Newton's equation of motion has a bigger invariance group than the Galilean group?
 A: Just to clarify before getting into the post, I prefer to avoid "prime" notation unless it is absolutely needed, so instead I will refer to "old coordinates" as $(x^{1}, x^{2}, x^{3})$ and "new coordinates" as $(y^{1}, y^{2}, y^{3})$. Hope that doesn't cause too much confusion. Also, I try to clarify at the end of my post why this post does not conflict with @Ryder Rude's post.

If we however consider the general coordinate transformation of the form $x'^i=x'^i(x^1,x^2,x^3)$, for example, going from Cartesian to spherical or cylindrical coordinates,  doesn't the form of Newton's also remain unchanged in the sense that if $F^i=m\frac{d^2x^i}{dt^2}$ holds in the Cartesian coordinates, ${F^i}^\prime=m\frac{d^2x^{\prime i}}{dt^2}$ holds in the Spherical or cylindrical coordinates, where $F^i$ and $\frac{d^2x^i}{dt^2}$ transform as $$F^{\prime i}=\frac{\partial x^{\prime i}}{\partial x^j}F^j, \quad \frac{d^2x^{\prime i}}{dt^2}=\frac{\partial x^{\prime i}}{\partial x^j}\frac{d^2x^j}{dt^2}.$$

This is false. Anything with a coordinate second time-derivative such as acceleration does not transform covariantly. This is surprising and subtle fact. The real transformation for acceleration is
$$ \frac{d^{2}x^{i}}{dt^{2}} = \frac{d}{dt}\left( \frac{dx^{i}}{dt} \right) = \frac{d}{dt}\left( \frac{\partial x^{i}}{\partial y^{j}}\frac{dy^{j}}{dt} \right) = \frac{d}{dt}\left( \frac{\partial x^{i}}{\partial y^{j}} \right)\frac{dy^{j}}{dt} + \frac{d^{2}y^{j}}{dt^{2}}\frac{\partial x^{i}}{\partial y^{j}} \ne \frac{d^{2}y^{j}}{dt^{2}}\frac{\partial x^{i}}{\partial y^{j}} $$
in the general case (that last inequality turns into an equality only in special cases). Force and velocity, on the other hand, are covariant. The framework that clarifies this apparent inconsistency is Riemannian geometry, although I won't go much into in it here.

Cylindrical Coordinates Example
Let me provide you with an explicit example. I'll provide details. Unfortunately, going through spherical coordinates is much too cumbersome and I have only so much time, so I will go through cylindrical coordinates instead.
The Cartesian coordinates are $(x^{1}, x^{2}, x^{3})$. The cylindrical coordinates are $(r, \theta, z) = (y^{1}, y^{2}, y^{3})$. The transformations are
\begin{align*}
x^{1} = r\cos\theta, \qquad x^{2} = r\sin\theta, \qquad x^{3} = z,
\end{align*}
and
\begin{align*}
r = \sqrt{(x^{1}) + (x^{2})}, \qquad \theta = \arctan(x^{2}/x^{1}), \qquad z = x^{3}.
\end{align*}
Now
\begin{align*}
\dot{x}^{1} &= \dot{r}\cos\theta - r\dot{\theta}\sin\theta, \\
\dot{x}^{2} &= \dot{r}\sin\theta + r\dot{\theta}\cos\theta, \\
\dot{x}^{3} &= \dot{z},
\end{align*}
and
\begin{align*}
\ddot{x}^{1} &= \ddot{r}\cos\theta - 2\dot{r}\dot{\theta}\sin\theta - r\ddot{\theta}\sin\theta - r\dot{\theta}^{2}\cos\theta, \\
\ddot{x}^{2} &= \ddot{r}\sin\theta + 2\dot{r}\dot{\theta}\cos\theta + r\ddot{\theta}\cos\theta - r\dot{\theta}^{2}\sin\theta, \\
\ddot{x}^{3} &= \ddot{z}.
\end{align*}
Next, we will look at how the force vector transforms, which is postulated to transform covariantly.
I will denote the $x$-components of the force as $F^{i}$ and the $y$-components of the force as $\overline{F}^{i}$. In this notation,
\begin{align*}
\overline{F}^{1} &= \frac{\partial y^{1}}{\partial x^{1}}F^{1} + \frac{\partial y^{1}}{\partial x^{2}}F^{2} + \frac{\partial y^{1}}{\partial x^{3}}F^{3}, \\[1.2ex]
\overline{F}^{2} &= \frac{\partial y^{2}}{\partial x^{1}}F^{1} + \frac{\partial y^{2}}{\partial x^{2}}F^{1} + \frac{\partial y^{2}}{\partial x^{3}}F^{3}, \\[1.2ex]
\overline{F}^{3} &= \frac{\partial y^{3}}{\partial x^{1}}F^{1} + \frac{\partial y^{3}}{\partial x^{2}}F^{2} + \frac{\partial y^{3}}{\partial x^{3}}F^{3}. 
\end{align*}
Some work shows
\begin{align*}
\frac{\partial y^{1}}{\partial x^{1}} &= \cos\theta  & \frac{\partial y^{1}}{\partial x^{2}} &= \sin\theta & \frac{\partial y^{1}}{\partial x^{3}} &= 0 \\[1.2ex]
\frac{\partial y^{2}}{\partial x^{1}} &= -\frac{\sin\theta}{r}  & \frac{\partial y^{2}}{\partial x^{2}} &= \frac{\cos\theta}{r} & \frac{\partial y^{2}}{\partial x^{3}} &= 0 \\[1.2ex]
\frac{\partial y^{3}}{\partial x^{1}} &= 0  & \frac{\partial y^{3}}{\partial x^{2}} &= 0 & \frac{\partial y^{3}}{\partial x^{3}} &= 1.
\end{align*}
Putting all this information together and then simplifying, we arrive at the following equations:
\begin{align*}
\overline{F}^{1} &= m(\ddot{r} - r\dot{\theta}^{2}), \\
\overline{F}^{2} &= m(\frac{2\dot{r}\dot{\theta}}{r} + \ddot{\theta}), \\
\overline{F}^{3} &= m\ddot{z}.
\end{align*}
As you can see, this is clearly not $\overline{F}^{i} = m\ddot{y}^{i}$, so it's not of the same form. Thus we conclude that the form of Newton's second law is not invariant under the passive transformation $\text{Cartesian}\rightarrow\text{cylindrical}$.
The same story holds for spherical coordinates, so the corresponding passive transformation also fails to preserve the form of Newton's second law.

Let me clarify one more thing. So far, I've shown that the form of Newton's second law changes under $\text{Cartesian}\rightarrow\text{cylindrical}$ assuming we employ the coordinate time-derivative in $F = m(d^{2}x/dt^{2})$.
However, the story is different if we employed the covariant time-derivative in $F = m(D\dot{x}/dt)$. This makes the form of the acceleration term covariant under all possible passive coordinate transformations. In this view, Newton's second law is covariant (form invariant) under all coordinate transformations.
Of course, this is all done with passive transformations in mind. Active transformations give a different point-of-view, and @Ryder Rude's post is relevant in that regard.
A: The literal answer to your question is "yes", but not for the reasons you give. Galilean transformations preserve orientation, and thus don't include reflections, but Newton's equations are invariant under reflection.
For cylindrical and spherical coordinates, the issue comes down to the Jacobian being non-constant. Take a Galilean transformation such a rotation. The three-dimensional Jacobian is going to be something like $$\begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0  \\ \sin(\theta) & \cos(\theta) &0 \\ 0 & 0 &1 \end{bmatrix}$$ where $\theta$ is a fixed parameter. But going from cylindrical coordinates to Euclidean, you're going to have $$\begin{bmatrix}\cos(\theta) & -r\sin(\theta) & 0  \\ \sin(\theta) & r\cos(\theta) &0 \\ 0 & 0 &1 \end{bmatrix}$$ where $r$ and $\theta$ are the cylindrical coordinates. Those non-constant terms mean that when you take the second derivative, you'll have motion whose acceleration is zero in one coordinate system, and non-zero in another, and thus there is a force (or a pseudo-force) in one, but not in the other.
For instance, if we consider just two-dimensional polar coordinates, if we have a nonzero $\frac{\delta \theta}{\delta t}$, but $\frac{\delta r}{\delta t}$ and the second derivatives are zero, then we'll see that when we convert to Cartesian co-ordinates, there's a force towards the center. Conversely, for a path that's straight in Cartesian coordinates, there will be an outward force in polar coordinates (centrifugal "force").
And if we then add a positive $\frac{\delta r}{\delta t}$, while keeping the constant angular velocity, then in the Cartesian co-ordinates, the linear velocity must be increasing (linear velocity is radius times angular velocity, and the radius is increasing). Conversely, for a path with constant velocity in Cartesian coordinates that has an outward component, the angular velocity will be decreasing, and so in polar co-ordinates there will be a force in the angular direction (Coriolis "force").
And there's also the question: If your logic were sound, couldn't you apply it to any locally diffeomorphic transformation?
A: I think you are mixing up active and passive transformations. Classical mechanics is a theory on a manifold. It can be formulated in a
diffeomorphism invariant manner. See this post.
Diffeomorphism invariance is not some big statement about the physical content of a theory. It only says that co-ordinates are meaningless labels to describe the underlying manifold.
Gelilean Transformations, however, can be interpreted as active transformations. Say, you perform some experiment with some initial conditions. And then you perform another experiment with the rotated initial conditions. Keep in mind that we haven't changed the co-ordinate system. We have changed the experiment. Now, rotational invariance of the laws means that the results of your new experiment will be the rotated version of the results of your earlier experiment.
A change from cartesian to spherical or cylindrical co-ordinates cannot be interpreted as an active transformation, as the co-ordinate systems are different (for example, the angle is a periodic dimension). So these transformations are describing the same experiment with just different labels.
A passive transformation always maps solutions of the EoM to solutions, because the functional form of the Lagrangian $L$ is allowed to change under a passive co-ordinate transformation to keep the action invariant. But an arbitrary active transformation need not map solutions of the EoM to solutions, as the functional form of the Lagrangian is not allowed to change under an active transformation. So the action may not be invariant under an active transformation.
Galilean transformations are of special interest because they are active transformations that do map solutions to solutions.
