Convolutions with changes to the argument
Let me show you how to manipulate the convolution integral.
Let
$$y(t)  = x(t) \star h(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau \tag{1} $$
Let $a &gt; 1$ that we have a compressed signal $y(at)$ ;
$$y(at)  = \int_{\tau=-\infty}^{\infty} x(\tau) h(at-\tau) d\tau \tag{2} $$
Substitude $\tau' = \tau/a$ so that $\tau = a \tau'$ and $d\tau = a \cdot d\tau'$ into Eq.2 :
\begin{align}
y(at)  &amp;= \int_{\tau'=-\infty}^{\infty} x(a \tau')~ h(at-a\tau') ~a ~d\tau' \tag{3}\\ \\
&amp;= a \int_{\tau'=-\infty}^{\infty} x(a \tau') ~ h(a(t-\tau')) ~d\tau' \tag{4}\\ \\
&amp;= a \int_{\tau=-\infty}^{\infty} x(a \tau) ~ h(a(t-\tau)) ~d\tau  = a \cdot \big( x(at) \star h(at) \big) \tag{5}
\end{align}
Eq.5 shows that the compressed output $y(at)$ is obtained by the convolution of the compressed input $x(at)$ and the compressed impulse response $h(at)$, and (also the result scaled by $a$)....
Note that convolution operator assumes an LTI (linear time-invariant) framework. However, the compressor is not an LTI system and it does not have an impulse response $h_c(t)$, so that $y(at) = y(t) \star h_c(t)$ can not be true.
Assume for a moment that the compressor is an LTI system with impulse response $h_c(t)$. Then we can consider the generation of the compressed output in two (serial) LTI stages:
$$ x(t) \longrightarrow \boxed{ h(t) } \longrightarrow \boxed{ h_c(t)} \longrightarrow y(at) \tag{6}$$
From LTI theory we know that the cascade of two LTI blocks is also LTI with an equivalent impulse response given by $h_e(t) = h(t) \star h_c(t)$ hence Eq.6 can be written as :
$$ y(at) = x(t) \star h_e(t)  = x(t) \star ( h(t) \star h_c(t)) = x(t) \star h(at)\tag{7}$$
Which is contradicting the result we have obtained in Eq.5, hence the assumption that the compressor could have been represented with an impulse reponse was wrong.
Another implication of this is also observed in the time-reversal of the output $y(t)$ :
$$y(-t) = x(-t) \star h(-t) \neq x(-t) \star h(t) \neq x(t) \star h(-t) $$
The reason of the unequal signs is again about reversal operation not being an LTI system.
On the other hand, if you want to apply an LTI operation on the output such as $y_d(t) = y(t-d)$ , then it can be simply be reflected in the convolution as:
$$ y(t-d) = y(t) \star \delta(t-d) = (x(t) \star h(t)) \star \delta(t-d) = x(t) \star h(t-d)$$...