Correspondence Between First Galois Cohomology and Semilinear Actions Up to Isomorphism

I've been stuck for a while on Exercise 1.9 of Bjorn Poonen's "Rational Points on Varieties". We start with $$L/K$$ a finite Galois extension with Galois group $$G$$, some $$r \in \mathbb{Z}_{\geq 0}$$, and for each 1-cochain $$\xi: G \rightarrow GL_r\left(L\right)$$, we define $$W_{\xi}$$ to be $$L^r$$ with equipped with the function $$G \times L^r \rightarrow L^r$$, $$\sigma \mapsto \xi_{\sigma}\left(\sigma w\right)$$.

In the first part of the question, we are asked to show that this function describes a semilinear $$G$$-action on $$W_{\xi}$$ iff $$\xi$$ is a 1-cocycle, i.e., $$\xi_{\sigma \tau} = \xi_\sigma {}^{\sigma}\xi_{\tau}$$. I have managed this.

The second part is to show that, for two $$1$$-cocycles $$\xi$$ and $$\xi'$$, the spaces $$W_\xi$$ and $$W_{\xi'}$$ are isomorphic as $$L$$-vector spaces with semilinear $$G$$-action iff $$\xi$$ and $$\xi'$$ are cohomologous. This is where I am stuck.

I am new to group cohomology, but I have studied other types of cohomology before and have read a few things on group cohomology in order to try and solve this problem. As I understand it, $$\xi$$ and $$\xi'$$ are cohomologous iff their difference is a $$1$$-coboundary, i.e., a principal crossed homomorphism. Translating the standard definitions into multiplicative language, I believe this amounts to showing that there is some $$f \in GL_r\left(L\right)$$ such that, for all $$w \in W = L^r$$ and all $$\sigma \in G$$, we have $$\xi'_{\sigma} \xi_{\sigma}^{-1}\left(w\right) = \left({}^{\sigma}f\right) f^{-1}\left(w\right),$$ i.e.,

$$$$\xi'_{\sigma} \xi_{\sigma}^{-1} = \left({}^{\sigma}f\right) f^{-1}.$$$$

Now, from what I can gather, $$W_{\xi}$$ and $$W_{\xi'}$$ are isomorphic as $$L$$-vector spaces with semilinear $$G$$-action iff there is $$g \in GL_r\left(L\right)$$ such that for all $$w \in W = L^r$$ and all $$\sigma \in G$$, we have $$g\left(\xi_\sigma \left(\sigma w\right)\right) = \xi'_{\sigma}\left(\sigma g\left(w\right)\right),$$ i.e., $$g\left(\xi_\sigma \left(\sigma w\right)\right) = \xi'_{\sigma}\left( {}^{\sigma}g\left(\sigma w\right)\right),$$ meaning precisely that

$$$$g \xi{_\sigma} = \xi'_{\sigma}{}^{\sigma}g.$$$$

My idea for tackling this is to show that we can manipulate the second expression until it is in the form of the first one.

From $$\xi$$ being a 1-cocycle, we can deduce that $$\xi_{id_G} = id_{GL_r\left(L\right)}$$ and then that $$\xi_{\sigma}^{-1} = {}^{\sigma}\xi_{\sigma^{-1}}$$, so, assuming the latter highlighted equation (involving $$g$$), we get $$\xi'_{\sigma} \xi_{\sigma}^{-1} = \xi'_{\sigma} {}^{\sigma}\xi_{\sigma^{-1}} = g\xi_\sigma {}^{\sigma}\left(g^{-1}\right) {}^{\sigma}\xi_{\sigma^{-1}} = g\xi_\sigma {}^{\sigma}\left(g^{-1}\xi_{\sigma^{-1}}\right),$$ but I can't work that into the form I want by a suitable choice of $$f$$. I have tried other manipulations, but with no success.

My conclusion is that I have probably misunderstood or mistranslated some concept. In particular, I fear I might have the wrong notion of isomorphism for $$L$$-vector spaces with semilinear $$G$$-action. The definition I have here is one I found for isomorphism of $$G$$-sets on Wikipedia. Initially, I had wondered whether an isomorphism should be a pair $$\left(\phi,h\right)$$ where $$\phi: G \rightarrow G$$ is an automorphism, $$h \in Gl_r\left(L\right)$$ and, for all $$w \in W$$ and $$\sigma \in G$$, $$h\left(\xi_\sigma \left(\sigma w\right)\right) = \xi'_{\phi\left(\sigma\right)}\left(\phi\left(\sigma\right)h\left(w\right)\right),$$

so in particular I get the first definition back if $$\phi$$ is the identity map on $$G$$, but I'm not sure.

If anybody can point out the erroneous definitions/where I am going wrong, then I would be very grateful.

Your mistake is in the meaning of $$\xi$$ and $$\xi'$$ being cohomologous. The condition should be $$\xi_\sigma =f^\sigma \xi'_\sigma f^{-1}$$, for some $$f\in GL_r(L)$$. This is the correct definition of cohomologous cocycles in the non-abelian cituation, and then $$f$$ just give you the isomorphism.