When in a vector space, say R3, I make a basis change from the orthonormal basis to some oblique group of three vectors for example, am I changing also the inner product and the definition of distance in that space?

No, changing basis doesn't change the inner product or redefine length. A key idea here is that the vectors exist whether you have a basis or not, and so the inner product is a relationship between vectors in the abstract. An inner product  fundamentally has nothing to do with any basis at all. For me, this is easiest to think about in the context of physics. Think of two vectors as floating around in space like this: We will define their inner product with the formula ⃗a⋅⃗b=abcos(θ)a→⋅b→=abcos⁡(θ)\vec{a}\cdot\vec{b} = ab \cos(\theta) On the left side, we see the inner product of two vectors. On the right, we have their lengths and the angle between them. Everything that goes into calculating the inner product is something you could physically measure with a ruler and a protractor and can be labeled directly on the picture. There is no coordinate system involved at all here. There's no basis chosen. Despite that, the idea of an inner product still makes sense as long as lengths and angles do. Our task will be to figure out how to make sure nothing gets screwed up once we do have a basis. Note that this definition has the properties of symmetry ⃗a⋅⃗b=⃗b⋅⃗aa→⋅b→=b→⋅a→\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a} and linearity ⃗aâ‹…(α⃗b+β⃗c)=α⃗a⋅⃗b+β⃗a⋅⃗ca→⋅(αb→+βc→)=αa→⋅b→+βa→⋅c→\vec{a}\cdot(\alpha\vec{b} + \beta\vec{c}) = \alpha \vec{a}\cdot\vec{b} + \beta \vec{a}\cdot\vec{c} that characterize an inner product. For our own convenience, we next add in a coordinate system, like this: The vectors can now be written in terms of coordinates and unit vectors, like this: ⃗a=ax^x+ay^ya→=axx^+ayy^\vec{a} = a_x\hat{x} + a_y\hat{y} ⃗b=bx^x+by^yb→=bxx^+byy^\vec{b} = b_x\hat{x} + b_y\hat{y} From the picture, it appears ax≈1.8ax≈1.8 a_x \approx 1.8 etc. Now that we have a coordinate system to work with, we can figure out what the inner product ought to be. We see that ^xâ‹…^x=^yâ‹…^y=1x^â‹…x^=y^â‹…y^=1\hat{x}\cdot\hat{x} = \hat{y}\cdot\hat{y} = 1 and that ^xâ‹…^y=0x^â‹…y^=0\hat{x}\cdot\hat{y} = 0. Thus we conclude This is the standard Euclidean inner product that you're used to. But what if we chose a different basis, like this: We can still say ⃗a=a′x^x′+a′y^y′a→=ax′x′^+ay′y′^\vec{a} = a'_x\hat{x'} + a'_y\hat{y'} ⃗b=b′x^x′+b′y^y′b→=bx′x′^+by′y′^\vec{b} = b'_x\hat{x'} + b'_y\hat{y'} Notice that the components and the basis vectors change, but the vectors themselves do not. They still represent that same physical objects. It's just the numbers that go with them that are different. In this case, it appears, for example, that ⃗a≈1.5^y′a→≈1.5y′^ \vec{a} \approx 1.5 \hat{y'} The inner product between the two vectors is unchanged because, as you recall, the inner product is something that depends on ruler and protractor measurements. Rulers and protractors don't change just because we picked a different basis. The line from before becomes We cannot simplify further until we know the inner products the unit vectors in this coordinate system. For example, let's take the example The inner product is 2*1 + 3*4 = 14. Let's choose a new basis ^x′=2^x−^yx′^=2x^−y^\hat{x'} = 2\hat{x} - \hat{y} ^y′=^x+^yy′^=x^+y^\hat{y'} = \hat{x} + \hat{y} You can verify that in this new basis, the vectors are now ⃗a=−13^x′+83^y′a→=−13x′^+83y′^\vec{a} = \frac{-1}{3} \hat{x'} + \frac{8}{3}\hat{y'} ⃗b=−^x′+3^y′b→=−x′^+3y′^\vec{b} = -\hat{x'} + 3\hat{y'} Furthermore, the inner products between the unit vectors are ^x′⋅^x′=5x′^â‹…x′^=5 \hat{x'} \cdot \hat{x'} = 5 ^x′⋅^y′=1x′^â‹…y′^=1\hat{x'} \cdot \hat{y'} = 1 ^y′⋅^y′=2y′^â‹…y′^=2\hat{y'}\cdot \hat{y'} = 2 That means the formula for the inner product in this basis is ⃗a⋅⃗b=5a′xb′x+a′xb′y+a′yb′x+2a′yb′ya→⋅b→=5ax′bx′+ax′by′+ay′bx′+2ay′by′\vec{a}\cdot\vec{b} = 5a'_xb'_x + a'_xb'_y + a'_yb'_x + 2a'_yb'_y The inner product computed in the primed basis, via plugging into the earlier formula, is This is the same thing we got in the original basis. The inner product doesn't depend on the basis. However, if you want to use the components of a vector, not the vector itself, to compute the inner product, your formula will change when you change basis. This viewpoint, where the vectors exist independently of the basis you're using, is helpful conceptually in physics. When we have an equation like ⃗F=m⃗aF→=ma→\vec{F} = m\vec{a} it is a relationship purely between vectors, and so it will be true whatever basis we use. Different bases correspond to difference frames of reference. So although we generally due physics calculations in one particular frame of reference, the results are true in all frames.