An Orthogonal Transformation of IID Standard Normal $X$ Is Also IID Standard Normal

Suppose that $X_1, X_2 ... X_n$ are IID standard normal variables and that $V$ is an orthogonal transformation. The text is claiming that $Y = VX$ is also IID standard normal because the two joint densities are the same:

$$\begin{split} P( Y \in \Omega) \ \ = \ \ P(X \in V^{-1}\Omega) \ \ & = \ \ \int_{V^{-1}\Omega} f(x) \, dx \\ & = \ \ \int_\Omega \frac{f(V^{-1}y)}{| \det V |} \, dy \ \ = \ \ \int_{\Omega} f(V^{-1} y) \, dy \\ \\ f(x) \ \ = \ \ f(V^{-1}y) \ \ & = \ \ (2 \pi)^{-n/2}e^{-|V^{-1}y|^2/2} \ \ = \ \ (2 \pi)^{-n/2} e^{-|y|^2/2} \ \ = \ \ f(y) \end{split}$$

However, each element in $Y$ is still a linear combination of the same $X_1, X_2 \ldots X_n$. It is in this respect that I'm having some trouble imagining that $Y_1, Y_2 \ldots Y_n$ are also IID standard normal. Is there another explanation to this (in my view) surprising result?

Note $X_1,...,X_n$ are iid standard normal iff they are jointly standard normal: $$X\equiv(X_1,...,X_n)'\sim N(0,\mathbb{I}_n).$$

Any affine transform of (multivariate) normal is also normal, so $VX$ is normal with mean $V\mathbb{E}[X]=0$ and variance $V\text{Var}(X)V'=VV'=\mathbb{I}_n$ by orthogonality of $V$, implying $(VX)_1,...,(VX)_n$ are iid standard normal.