Linear vector transformations in 3Blue1Brown with basis vectors

I am looking at the 3Blue1Brown vector transformation video and at 2:17 he gives an example of a transformation that rotates space 90 degrees over the y-axis.

Based on the intuition I have it seems weird because he says that $$\hat{i}, \hat{j}$$ and $$\hat{k}$$ are all basis vectors. Therefore in a 3-dimensional Euclidean space the vectors should have the values $$(1,0,0), (0,1,0)$$ and $$(0,0,1)$$ respectively.

Now I have the understanding that every transformation can be expressed as some scalar multiplicaton of the basis vectors. For exaple if I have a vector $$\overrightarrow{v} = (2,3)$$ where the $$2$$ and $$3$$ are the coordinates of the vectors that can be represented in relation to the basis vectors: $$\overrightarrow{v} = (2\hat{i}, 3\hat{j})$$ on a two dimensional plane and want to rotate it 180 degrees then the transformation could also be expressed as multiplying each coordinate of the vector with a scalar: $$(-1\cdot \overrightarrow{v}, -1\cdot \overrightarrow{v})$$ or using matrices $$\begin{pmatrix}2\\ 3\end{pmatrix} \begin{pmatrix} -1, -1 \end{pmatrix}$$.

That said in the video where he performs the transformation visually he shows how the $$\hat{i}$$ changes from the x-axis to the z-axis which is weird because to perform some transformation changing axies should not be neccessary and in my intuition cannot be done because we have defined $$\hat{i} = (1,0,0)$$

My question is that if we assign concrete values for for example $$\hat{i}$$ then we shouldnt be able to transform them from one axis to the other, all basis vectors represent one axis but in the video the basis vectors change axies

Let's consider $$2$$ dimensions first. When we perform a rotation of the plane by $$90º$$ counterclockwise, what we're really doing is moving $$\hat{i}$$ to $$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$$ and $$\hat{j}$$ to $$\begin{pmatrix} -1 \\ 0\end{pmatrix}$$. In other words, $$\hat{i}$$ is now pointing in the positive $$y$$- direction, and $$\hat{j}$$ is now pointing in the negative $$x$$-direction.
Because addition is preserved under a linear transformation (transforming the sum of two vectors is the same as transforming each separately then adding them together), and because scalar multiplications are preserved ($$f(u + v) = f(u) + f(v)$$ and $$f(cu) = cf(u)$$ more formally), we can decompose every transformation in terms of these basis vectors.
So in the case of $$3$$ dimensions, or $$n$$ dimensions, we can move the basis vectors anywhere, even to another axis, and any transformation can be thought of in this way.