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Let $M$ and $N$ be Riemannian manifolds. My understanding is that strictly speaking, a diffeomorphism $\phi:M \to N$ only acts on the smooth manifold structure, not the metric tensor. But there is a natural action of $\phi$ on the metric tensor given by the pullback under the diffeomorphism, which in fact makes the diffeomorphism an isometry. If no diffeomorphism between $M$ and $N$ is an isometry, is it standard to still refer to the Riemannian manifolds as "diffeomorphic"?

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  • $\begingroup$ When you say "if no diffeomorphism between $M$ and $N$ is an isometry", do you mean $M$ with its original metric, or with the pullback metric defined previously? With the pullback, there's an isometry by definition. With the original... consider two spheres of different size. $\endgroup$ – user357151 Aug 19 '18 at 20:50
  • $\begingroup$ Sadly, the question in your title and the question in your last sentence are at odds with one another. :) $\endgroup$ – Ted Shifrin Aug 19 '18 at 23:51
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The answer to your last question ("If no diffeomorphism between $M$ and $N$ is an isometry, is it standard to still refer to the Riemannian manifolds as "diffeomorphic"?) is Yes. (As @TedShifrin pointed out, this is at odds with the question in your title, to which the answer is No.)

To say that two smooth manifolds are diffeomorphic means there exists a diffeomorphism between them, irrespective of what that diffeomorphism does to any metric. If the manifolds are also endowed with Riemannian metrics, we say that they are isometric if there exists a diffeomorphism between them that pulls one metric back to the other.

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    $\begingroup$ Oh, good point, @TedShifrin . I've edited my answer to make it clearer which question I was answering "yes" to. $\endgroup$ – Jack Lee Aug 19 '18 at 23:55

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