We know that the spaces $L^p(\Omega)$ and $L^q(\Omega)$ are isometric and isomorphic for $p,q$ conjugate and $p,q \neq 1,\infty$. Call the isomorphism $l\colon L^p(\Omega) \to L^q(\Omega)$.
Take $p=q=2$. $L^2(\Omega)$ is a Hilbert space so we have a isometric isomorphism $r$ between $L^2(\Omega)$ and $L^2(\Omega)^*$ which is the Riesz map. Is this the same map as the above? I.e. is $l \equiv r$?
Are you sure that we know that $L^p(\Omega)$ is isometrically isomorphic to $L^q(\Omega)$ for $p,q$ conjugate exponents? We actually know that $L^p$ is isometrically isomorphic to the dual of $L^q$, not to $L^q$ itself. An immediate example to show that this is not true: consider the measure space $\mathbb{N}$ with the counting measure, so the $L^p$ spaces are actually the $\ell^p$ spaces. we know that for $0<s<t<\infty$ it is $\ell^s\subset\ell^t$. Indeed, if $(x_n)\in \ell^s$, then $\sum_n|x_n|^s<\infty$ so $|x_n|^s$ becomes less than $1$ after a certain point, so $|x_n|$ becomes less than 1 after a certain point, so,since $s<t$ it will be $|x_n|^t\leq|x_n|^t$ from that point on which shows that $(x_n)\in\ell^t$. Now take for example the conjugate exponents $p=3/2$ and $q=3$. We have that $\ell^{3/2}\subset\ell^3$, but this inclusion is not isometric: for example, the sequence $x=(\frac{1}{2^n})_{n=1}^\infty$ satisfies $$\|x\|_{3/2}=\sum_{n=1}^\infty\frac{1}{2^{3n/2}}=\frac{2^{-3/2}}{1-2^{-3/2}} $$ while on the other hand it is $$\|x\|_3=\sum_{n=1}^\infty\frac{1}{2^{3n}}=\frac{8^{-1}}{1-8^{-1}}$$
Any Hilbert space is isometrically isomorphic to its dual via the Riesz map $$H\to H^*,\;\;\;x\mapsto\langle-,x\rangle $$
The isometric isomorphism between $L^p$ and $(L^q)^*$ is given by $f\mapsto\varphi_f:L^q\to\mathbb{C}$ where $\varphi_f(g)=\int\overline{f}g$. Since the inner product on $L^2$ is $\langle f,g\rangle=\int f\overline{g}$, we actually see that the Riesz map of $L^2$ coincides with the isometric isomorphism between $L^2$ and $(L^2)^*$.