I am looking for a reference that proves using the (in)homogeneous bar complex for group cohomology that the $\mathrm{Aut}(G)$ action on $H^\bullet(G,\mathbb{R})$ factors through $\mathrm{Out}(G)$. All the proofs I have found so far use a topological or derived functors perspective on group cohomology, which I would like to avoid. (I do not have enough space.)
I know that the group cohomology is the cohomology of the homogeneous bar complex:
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} % \begin{array}{llllllllllll} 0 & \ra{} & \mathrm{Map}(G,\mathbb{R})^G & \ra{d^0} & \mathrm{Map}(G^2,\mathbb{R})^G & \ra{d^1} & \mathrm{Map}(G^3,\mathbb{R})^G & \ra{d^2} &... \end{array} $$ where \begin{equation} d^n f(g_0,...,g_{n+1})=\sum_{i=0}^{n+1} (-1)^i f(g_0,...,\hat{g_i},...,g_{n+1}). \end{equation}
Then $\mathrm{Aut}(G)\curvearrowright \mathrm{Map}(G^{n+1},\mathbb{R})^G$ by $\sigma.f(g_0,...,g_n)=f(\sigma^{-1}g_0,...,\sigma^{-1}g_n)$. I understand why this gives an action $\mathrm{Aut}(G)\curvearrowright H^\bullet(G,\mathbb{R})$. Now to prove that this factors through $\mathrm{Out}(G)\curvearrowright H^\bullet(G,\mathbb{R})$ I have to find for every $h\in G$ and $f_1\in \mathrm{Map}(G^{n+1},\mathbb{R})^G$ an element $f_2\in \mathrm{Map}(G^{n},\mathbb{R})^G$ such that $f_1(hg_0h^{-1},...,hg_nh^{-1})-f_1(g_0,...,g_n)=d^{n-1}f_2$. This is the point, where I get stuck. What is $f_2$?
I would be equally happy for a proof using the inhomogeneous bar complex:$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} % \begin{array}{llllllllllll} 0 & \ra{} & \mathrm{Map}(G^0,\mathbb{R}) & \ra{\partial^0} & \mathrm{Map}(G^1,\mathbb{R}) & \ra{\partial^1} & \mathrm{Map}(G^2,\mathbb{R}) & \ra{\partial^2} &... \end{array} $$ where \begin{equation}\begin{split} \partial^n f(g_0,...,g_n)=\ &g_0f(g_1,...,g_n)+(-1)^{n+1}f(g_0,...,g_{n-1})\\ &-\sum_{j=0}^{n-1} (-1)^{j+1} f(g_0,...,g_jg_{j+1},...,g_n). \end{split}\end{equation}
Added: I know a proof that conjugation acts trivially on cohomology using syzygys, dimension shifting. But then I do not know why this is the same action as the one given on the bar complexes. Also I thought there might be an easier way?