Beside the algebraic definition of isometry:
Question 1. Is it correct that the isometry group of a Riemannian manifold depends on its embedding? e.g. I think isometry group of a circle in plan and in 3-space are different?
Question 2. Why we can talk about trivial isometry group while always a translation is a nontrivial member of isometry group?
After these, I think it would be clear for me that why an isometry of a Riemannian manifold, also called a symmetry. (The strange point for me is: members of isometry group like translation)
Best Answer
OK, let me see if I can answer your two questions.
But first, let's clear up a common misunderstanding. In Riemannian geometry, the word "isometry" is used with two different meanings. First, if $(M_1,g_1)$ and $(M_2,g_2)$ are Riemannian manifolds, an isometry from $\boldsymbol {M_1}$ to $\boldsymbol {M_2}$ is a smooth map $F\colon M_1\to M_2$ that satisfies $F^*g_2 = g_1$. For a given pair of Riemannian manifolds, there may or may not be any isometries between them.
Second, if $(M,g)$ is a fixed Riemannian manifold, an isometry of $\boldsymbol M$ is an isometry from $(M,g)$ to itself, that is, a smooth map $F\colon M\to M$ such that $F^*g = g$. In this case, the set of isometries of $M$ is a group under composition. It always contains at least the identity map, and it might or might not contain others. If $M=\mathbb R^{n+1}$ with its Euclidean metric, the isometry group contains all translations, rotations, reflections, and glide reflections. If $M=\mathbb S^n$ with its standard round metric, the group contains only (restrictions of) rotations and reflections.
To address your questions:
Question 1: The isometry group of a given Riemannian manifold $(M,g)$ depends only on the manifold $M$ and the metric $g$, not on any particular isometric embedding into a Euclidean space. This is immediate from the definition I gave above. (Of course, if you choose an embedding that is not an isometric embedding, then you will induce a different metric on $M$, and it will very likely have a different group of isometries.)
Question 2: The term "translation" only makes sense if we're talking about a vector space (or more generally an affine space). All translations of $\mathbb R^{n+1}$ are isometries of its standard metric, but there are also metrics on $\mathbb R^{n+1}$ that are not translation-invariant. You can't talk about "translations of $\mathbb S^n$" because there are no translations that take $\mathbb S^n$ to itself.
Of course, any translation of $\mathbb R^{n+1}$ takes $\mathbb S^n$ to another unit sphere, let's call it $S'$, and the restriction of that translation becomes an isometry from $\mathbb S^n$ to $S'$. But that's not an isometry of $\mathbb S^n$.