I'm reading the Wikipedia article on Pontryagin duality.
(This is a fuzzy question about intuition for the difference between the forward and inverse Fourier transforms in this context.)
It defines the Fourier transform for all $\chi$ in $\widehat{G}$: $$\widehat f(\chi) = \int_G f(x) \overline{\chi(x)}\ d\mu(x)$$
And the inverse transform $$ \check{g} (x) = \int_{\widehat{G}} g(\chi) \chi(x)\ d\nu(\chi)$$
The transform on $\widehat{G}$ is given by: $$ \begin{aligned} \widehat{g}(\operatorname{ev}_G(x)) &= \int g(\chi) \overline { \operatorname{ev}_G(x)(\chi) } d\nu(\chi) \\ &= \int g(\chi) \overline { \chi(x) } d\nu(\chi) \end{aligned} $$ where $\operatorname{ev}_G(x)$ for $x$ in $G$ is a generic element of $\widehat{\widehat{G}}$.
What is the intuitive reason for the minus sign in $$ \check{g}(x) = \widehat{g} ( \operatorname{ev}_G(-x)) $$
Is the isomorphism $\psi : G \to \widehat{\widehat{G}}$ given by $x \mapsto \operatorname{ev}_G(-x)$ more "natural" than $\operatorname{ev}_G$?
The Fourier inversion theorem is saying that the Fourier transform $W: L^2(G)\to L^2(\hat{G})$ is unitary $$\|W(f)\|_{L^2(\hat{G})} = \|f\|_{L^2(G)}$$ so that its inverse is its adjoint $W^*$, which is defined by $$\langle Wf,\hat{h} \rangle=\langle f,W^*\hat{h} \rangle,\qquad \forall f\in L^2(G),\hat{h}\in L^2(\hat{G})$$ When everything is $L^1$ we can change the order of integration to get
$$\int_G f(x)\overline{\int_{\hat{G}} \hat{h}(\chi) \chi(x)}d\chi dx=\int_{\hat{G}} \overline{\hat{h}(\chi)} \int_G f(x)\overline{\chi(x)}dx d\chi = \langle Wf,\hat{h} \rangle$$ whence $W^* \hat{h}(x) = \int_{\hat{G}} \hat{h}(\chi) \chi(x)d\chi$.
That $(WW f)\circ ev_G(x)= f(-x)$ is then a 'coincidence' coming from $G=\hat{\hat{G}}\circ ev_G$ and $\overline{\psi(t)}=\psi(-t)$.
That said proving the Pontryagin Fourier inversion theorem seems the only way to really see what is happening.