HKR generalized character theory question regarding a certain construction In that paper https://web.math.rochester.edu/people/faculty/doug/mypapers/hkr.pdf Hopkins-Kuhn-Ravenel introduced the idea of generalized character corresponding to a complex-oriented cohomology theory $E^*$. They construct an $E^*$-algebra $L(E^*)$, which is defined to be the colimit: ${\rm colim} E^*(B\mathbf{Z}^n_p)$ (More details regarding the whole construction can be found in the link.)
Later on, they use this construction to define the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$, where ${\rm Aut}(\mathbf{Z}^n_p)$ acts as $E^*$-algebra homomorphisms ($\mathbf{Z}_p$ denotes the additive group of $p$-adic integers). To prove this they define $L_r(E^*)=E^*(B\mathbf{Z}_{p^r})$, and the natural ${\rm Aut}(\mathbf{Z}_{p^r})$ action on gives $L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$. 
My question has to do with the proof of the above: What I understand is that the invariant rings $L_r(E^*)^{{\rm Aut}(\mathbf{Z}_{p^r})}=p^{-1}E^*$, induce a direct system of $E^*$-algebras and the colimit must be $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$. However, they don't give any proof hence should be somehow straightforward why the colimit is the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$ (not at all to me). Can you explain me please if my understanding makes sense? If yes, probably an explanation why the above colimit converges on the invariant ring $L(E^*)^{{\rm Aut}(\mathbf{Z}^n_p)}$ would be really helpful. If not, a sort of insight would be very appreciable!
 A: I think what you understand is correct.  Write $A_r= L_r(E^*)$, which fit into a direct system $A_r\to A_{r+1}\to \cdots$.  Let $G=\mathrm{Aut}(\mathbb{Z}_p^n)$, which acts compatibly on every $A_r$ (it actually acts on $A_r$ through the quotient group $\mathrm{Aut}((\mathbb{Z}/p^r)^n)$).  
If $\cdots \to A_r\to A_{r+1}\to \cdots$ is a sequence of injective maps between $G$-sets, then it is straightforward to show that $\mathrm{colim} (A_r^G) \to (\mathrm{colim} A_r)^G$ is an isomoprhism.  In fact, you only need them to be injective for all $r\geq R$ for some $R$. 
It remains to show that the $A_r\to A_{r+1}$ are injective.  I don't see a proof of this in the paper, but it is surely true.
Here's a proof that $A_r\to A_{r+1}$ is injective if $r\geq1$.  There's probably a better proof, but it's what comes to mind now.  I use the description in the proof of [HKR, 6.5]:
$$
A_r = S_r^{-1} E^*B\Lambda_r,
$$
where $\Lambda_n=(\mathbb Z/p^r)^n$, and $S_r\subseteq E^*B\Lambda_r$ is a certain multiplicatively closed subset, which can be defined as follows: 
$$
S_r = \{ c(\alpha):=(B\alpha)^*(x)\; | \; \alpha\in \Lambda_r^*\smallsetminus\{0\} \},\qquad \Lambda_r^*:=\mathrm{Hom}(\Lambda_r, U(1)),
$$
where $x\in E^*BU(1)$ is a chosen coordinate of the formal group.  Although the set $S_r$ depends on the choice of $x$, the fraction ring $A_r$ doesn't, because any two coordinates differ by a unit in $E^0BU(1)$.  Note that the direct system of $A_r$ comes from the inverse system $\cdots\to\Lambda_{r+1}\to \Lambda_r\to\cdots$
Note that $c(\alpha^p)=[p](c(\alpha))= c(\alpha)f(c(\alpha))$ where $[p](x)$ is the $p$-series of the formal group and $f(x)=[p](x)/x$ is a power series. This implies that if we invert $c(\alpha^p)$ then we automatically invert $c(\alpha)$ as well.  For  every $\alpha\in \Lambda_r^*\smallsetminus\{0\}$ with $r\geq1$, some $\alpha^{p^k}$ is in the image of  $\Lambda_1^*\to \Lambda_r^*$, so in fact 
$$
A_r = S_1^{-1}E^*B\Lambda_r,
$$
where $S_1$ really means the image of $S_1\subset E^*B\Lambda_1$ under the map induced by $\Lambda_r\to \Lambda_1$.  Since $E^*B\Lambda_r\to E^*B\Lambda_{r+1}$ is injective and $S_1^{-1}E^*B\Lambda_1$ is flat over $E^*B\Lambda_1$, the claim follows.
