I only have seen that for calculating the adjoint of an operator in $\mathcal S$, it used integration by parts, but I was thinking that if one can use substitution to find the adjoint. For exmple, for $f,g\in\mathcal S$
$$\langle f(ax),g(x)\rangle=\int_{-\infty}^{\infty}{f(ax)\ g(x)\ dx}$$
$$=\int_{-\infty}^{\infty}{f(u)\ g\left({u\over a}\right)\ {dx\over |a|}}={1\over |a|}\left\langle f(u), g\left({u\over a}\right)\right\rangle$$
using the substitution $u:=ax$.
So if we define an operator $\hat{T}_a:\mathcal S\to \mathcal S$, $a\in \Bbb R$, that maps $f(x)\mapsto f(ax)$, the adjoint of $\hat{T}_a$ should be ${1\over |a|}\hat{T}_{1/a}$?
For another example: the Lagrange shift operator ($L_t$) by $t$-units, where $t\in \Bbb R$, it can be demostrated using integration by parts, and then the linearity of inner product apply to each term of the polynomial $\exp(t\partial_x)$; that the adjoint of $T_t$ is $T_{-t}$. That could come instead of using integration by parts, just using the substitution $u:=x+t$