If $G=p^n$, how does the action of $\operatorname{Aut}(G)$ on $G/G'G^p$ induce a homomorphism $\operatorname{Aut}(G)\to GL_m(\mathbb{F}_p)$?

Question

Say $G$ is finite $p$-group of order $p^n$. I know that $G/G'G^p$ is also a $p$-group, and let's say it has order $p^m$. I can define an action of $\operatorname{Aut}(G)$ on $G/G'G^p$ via $f*gG'G^p:=f(g)G'G^p$. I am confused as to how this action gives us a homomorphism $\operatorname{Aut}(G) \to GL_m(\mathbb{F}_p)$. In particular, I am confused how the homomorphism will be defined. I can't really see how the action is meant to be used to make the $m\times m$ matrices over $\mathbb{F}_p$.