Define $S:= \{\sqrt{2}\sin(\pi n x): n\in\mathbb{N}_0\}$ where $N_0 = \{0,1,2,\dots\}$. I want to show that $S$ is an orthonormal basis for the space $L^2([0,1])$ when it is know that the set $\tilde{S}:= \{e^{2\pi il x}:l\in\mathbb{Z}\}$ forms an ONB for $L^2([0,1])$. I have been given a hint to construct a suitable isomorphism between $S$ and $\tilde{S}$, but I am not really sure how. I initially thought of mapping each $\sqrt{2}\sin(\pi n x)$ to some $e^{2\pi il x}$ by dividing with $\sqrt{2}$, multiplying the argument $x$ by two and introducing a sign depending on parity of $n$, i.e.
$$\sqrt{2}\sin(\pi n x)) \mapsto \sin(2\pi f(n)x + \pi/2) + i\sin(2\pi f(n)x)$$
where $f$ is a bijection from $\mathbb{N}_0$ to $\mathbb{Z}$. Under this mapping inner product should be preserved by orthogonality of the sines. My problem is that I don't know how to extend this map to the whole space $L^2([0,1])$.
How close is this construction to a proper solution and is there an easier way to achieve the claim?
If one extends any function $f\in L_p[0,1]$ to a function $f_o$ on $[-1,1]$ so that $f_o=(-x)=-f(x)$ for $0\leq x\leq 1$, then the extension $f_o$ is an odd function and $f_o\in L_p(-1,1)$ since $$\|f_o\|_{L_p[-1,1]}=2^{1/p}\|f\|_{L_p[0,1]}$$
Now,for the case $p=2$. The Fourier series (in terms of $f_o$ as an element of $L_2[-1,1]$) contains only sine terms ($f_o$ being odd). This will show that $\{S_n(x)=\sin(\pi n x):n\in\mathbb{N}\}$ is indeed an orthogonal basis for $L^2[0,1]$.
The linear map such that $S_n\mapsto \frac{e_n-e_{-n}}{i\sqrt{2}}\in L_2[-1,1]$, where $e_n(x)=e^{\pi n ix}$,$n\in\mathbb{N}$ defines the isometry the OP is looking for.
Similar approach can be used to show that $\{C_n(x)=\cos(\pi n x): n\in\mathbb{Z}_+\}$ is an orthogonal basis for $L_2[0,1]$. The extension of function $f$ on $[0,1]$ to a function $f_e$ on $[-1,1]$ is now set to be $f_e(-x)=f(x)$ for all $0\leq x\leq 1$.