Given two smooth manifolds with Riemannian metric $(X,g)$ and $(Y,h)$ and a smooth map $f: X \to Y$ I understand that we define $\phi$ to be an isometry if $f^* g = h$.
I thought I understood this definition but when trying to calculate it with an easy example (case where the manifolds are the same) I'm having difficulty - it should be straightforward and I'm probably missing something obvious!
I'm looking at a 1-parameter family of maps $f_t: S^2 \to S^2$ with $f_t(\theta,\phi) = (\theta + t, \phi)$ in local (spherical polar coordinates) where the spheres are given the same spherical polar metric $g = d\theta^2 + \sin^2(\theta)d\phi^2$.
Now I understand the condition for this to be an isometry is $f_t^*g(f_t(\theta,\phi)) = g(\theta,\phi)$ but we have $g(f_t(\theta,\phi)) = d\theta^2 + \sin^2(\theta+t)d\phi^2$ and then pulling this back (using the fact that $f^* (\omega dx) = f(w)d(f(w))$ generally) gives $f_t^*g(f_t(\theta,\phi)) = d\theta^2 + \sin^2(\theta+2t)d\phi^2$
It would be very helpful to understand where I am going wrong here.
First a small note: The condition for an isometry is $f^{*}h = g$; metrics pull back under smooth mappings.
The point with your example is, if you write the round metric $$ g = d\theta^{2} + \sin^{2}\theta\, d\phi^{2}, $$ you're using physicists' convention for the angle names: $\theta$ is colatitude and $\phi$ is longitude. The proposed coordinate translation is not, therefore, an isometry. The rotation mapping you presumably intended, $$ f_{t}(\theta, \phi) = (\theta, \phi + t), $$ is easily checked to be an isometry for each $t$.