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How to show that if $(S^1,g)$ and $(S^1,h)$ are of the same volume, then they are isometric? Here $g,h$ are Riemannian metrics on the sphere. How to explicitly construct an isometry?

So by the hint of Mariano Suárez-Álvarez, first define $l_g(t) = \int_0^t\langle r'(t), r'(t) \rangle_g^{\frac{1}{2}}dt$ with respect to the Riemannian metric $g$, where $r(t) = (cost,sint)$, in this case $l_g(t) = \int_0^t\langle \frac{\partial}{\partial x}|_{r(t)}, \frac{\partial}{\partial x}|_{r(t)}\rangle_g^{\frac{1}{2}}dt$. Now since $l(t)$ is bijective from $(0, 2\pi)$ to $(0,b)$ for some number $b$, then define $t_g(s) = l_g^{-1}$, so having the same volume, both $g,h$ will attain the same number $b$.

Now suppose that $t_g(b)=m \in S^1$ and $t_h(b)=n \in S^1$, then define $F:S^1 \to S^1$ in the following way: $\forall x \in S^1$, $\exists p \in (0,2\pi) $such that $r(p) = x$, then $F(x) = r(t_h(l(r^{-1}(x))))$. Now claim that $F$ is an isometry.

Is this right?

The idea as I see is to send points with the same arc length to the points with the same arc length. This is intuitive since to preserve the inner product, the length of the tangent vector at a point is sort of the limit length of the arc length at that point, so if arc length is preserved then I am done. Am I right?

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    It's not in general true that same volume implies isometric. For example, consider a round sphere, vs a(n american) football shape. By scaling the football shape, one can make the volume match that of the round sphere. But there two are not isometric as the round sphere has constant curvature by the football shape doesn't. – 2017-01-11
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    I am so sorry, and I have edited the question – 2017-01-11
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    The volume of a curve is its length. Parametrize the two curves by arc length. – 2017-01-11
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    Can you be more specific? @MarianoSuárez-Álvarez – 2017-01-11
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    I was very specific: you can parametrize curves by arc length. Do it for each of the two curves. Then think about what you've got. – 2017-01-11
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    I am not familiar with parametrizing curves by arc length, but I have attempted it in the edit. – 2017-01-11
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    Related: [Is there a smooth distance-persaving bijection between two simple curves with same length?](http://math.stackexchange.com/questions/1550128) – 2017-01-11
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    I do not quite understand your answer, why $R/l$ is isometric to $M$? How should I define an isometry explicitly in my case between two spheres? – 2017-01-11
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    If you do not know what parametrizing a curve by arc length is I empathically suggest that you spend some time reading a textbook on differential geometry of curves and surfaces! (It is somewhat weird that one can get to the point where one knows what the volume of a Riemannian manifold is and not that :-1 ) – 2017-01-11

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