Let $F^A$ be a curvature two form on a principal bundle $P$ corresponding to the connection one form $A$. Let $f:P\rightarrow P$ be a global bundle automorphism; there then exists a map $\sigma_f:P\rightarrow G$ such that:
$$f(p)=p\cdot \sigma_f(p)$$
Finally let $\mu_G$ be the maurer cartan form, given by:
$$\mu_G(X_g)=L_{g^{-1}*}X_g$$
where $X_g$ is a vector field on some Lie group $G$. Under the pullback of $A$ by $f$ we have:"
$$f^*A=\text{Ad}_{\sigma_{f^{-1}} } \circ A+\sigma_f^*\mu_G$$
I am trying to show that:
$$F^{f^*A}=\text{Ad}_{\sigma_{f^{-1}}}\circ F^A$$
where:
$$F^A=dA+\frac{1}{2}[A,A]$$
and for two Lie algebra valued one forms $\eta,\omega$:
$$[\eta,\omega](X,Y)=[\eta(X),\omega(Y)]-[\eta(Y),\omega(X)]$$
which comes from the definition of the wedge product for Lie algebra valued $k$ and $l$
forms.
When replacing $A$ with $f^*A$, we see that:
$$F^{f^*A}=d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)+\sigma_f^*d\mu_G+\frac{1}{2}\text{Ad}_{\sigma_{f^{-1}}}\circ [A,A]+\frac{1}{2}\sigma_f^*[\mu_G,\mu_G]+\frac{1}{2}[\text{Ad}_{\sigma_{f^{-1}}}\circ A,\sigma_f^*\mu_G]+\frac{1}{2}[\sigma_f^*\mu_G,\text{Ad}_{\sigma_{f^{-1}}}\circ A]$$
By the Maurer-Cartan equation we have that:
$$\sigma_f^*d\mu_G=-\frac{1}{2}\sigma^*_f[\mu_G,\mu_G]$$
hence:
$$F^{f^*A}=d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)+\frac{1}{2}\text{Ad}_{\sigma_{f^{-1}}}\circ [A,A]+\frac{1}{2}[\text{Ad}_{\sigma_{f^{-1}}}\circ A,\sigma_f^*\mu_G]+\frac{1}{2}[\sigma_f^*\mu_G,\text{Ad}_{\sigma_{f^{-1}}}\circ A]$$
My problem is now that I am unsure of how to calculate the exterior derivative of $\text{Ad}_{\sigma_{f^{-1}}}\circ A$. By our previous formula for two Lie algebra valued one forms we have that for $\eta=\text{Ad}_{\sigma_{f^{-1}}}\circ A$,
and $\omega=\sigma_f^*\mu_G$:
$$\frac{1}{2}\left([\eta,\omega]+[\omega,\eta]\right)(X,Y)=\frac{1}{2}\left([\eta(X),\omega(Y)]-[\eta(Y),\omega(X)]+[\omega(X),\eta(Y)]-[\omega(Y),\eta(X)]\right)$$
$$=[\eta(X),\omega(Y)]-[\eta(Y),\omega(X]$$
$$=[\eta,\omega](X,Y)$$
So I feel that:
$$d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)=\text{Ad}_{\sigma_{f^{-1}}}\circ dA-[\eta,\omega]$$
but I don't see why this should be true. In fact, since $\text{Ad}$ is a representation of $G$ on $\mathfrak{g}$ I would think that:
$$d(\text{Ad}_{\sigma_{f^{-1}}}\circ A)=\text{Ad}_{\sigma_{f^{-1}}}\circ dA$$
but then I have this extra factor of $[\eta,\omega]$, so I am at a loss. Any hints on how to calculate this exterior derivative, or on how that wedge product may actually be zero would be greatly beneficial.
Edit:
Employing Professor Shifrin's notation, I write that:
$$f^*A=\sigma_{f^{-1}}A\sigma_f+\sigma_{f^{-1}}d\sigma_f$$
I have already calculated the exterior derivative of the second term, albeit in a different notation, so I move to the first term:
$$d(\sigma_{f^{-1}}A\sigma_f)=d(\sigma_{f^{-1}})\wedge A\sigma_f+
\text{Ad}_{\sigma_f^{-1}}\circ dA+\sigma_{f^{-1}}A\wedge d(\sigma_f) $$
Note that:
$$\sigma_{f^{-1}}\sigma_f=e$$
hence:
$$d(\sigma_{f^{-1}})\sigma_f=-\sigma_{f^{-1}}d\sigma_f\Rightarrow d\sigma_{f^{-1}}=-\sigma_f^{-1}d\sigma_f\sigma_{f^{-1}}$$
Therefore we obtain:
$$d(\sigma_{f^{-1}}A\sigma_f)=-\sigma_{f^{-1}}d\sigma_f\wedge \text{Ad}_{\sigma_f^{-1}}A+
\text{Ad}_{\sigma_f^{-1}}\circ dA-\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge d(\sigma_{f^{-1}})\sigma_f$$
The final term in this sum can be rewritten as:
$$\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge d(\sigma_{f^{-1}})\sigma_f=-\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge \sigma_f^{-1}d\sigma_{f}$$
Thus:
$$d(\sigma_{f^{-1}}A\sigma_f)=-\sigma_{f^{-1}}d\sigma_f\wedge \text{Ad}_{\sigma_f^{-1}}A+
\text{Ad}_{\sigma_f^{-1}}\circ dA+\text{Ad}_{\sigma_{f^{-1}}}\circ A \wedge \sigma_f^{-1}d\sigma_{f}$$
From here, I can kind of see how the pieces should move, but I am unsure why. Reconciling the two notations is proving troublesome to me. To be clearer, it seems that from the definition for the wedge product of twisted forms, this product is actually symmetric for one forms, so I feel like the first term and the last term would cancel, which doesn't make sense. I am then left to conclude that there is something different about the wedge product I have used in this edit, and the wedge product I used in the original post, but I am unsure what that difference actually is, or how to make it precise.
Edit 2:
The definition of wedge product for Lie algebra valued one forms is
as follows:
Given a $\omega\in \Omega^k(P,\mathfrak{g})$ and $\eta\in \Omega^k(P,\mathfrak{g})$, then their wedge product is given by:
$$[\omega,\eta](X_1,\dots,X_{k+l})=\frac{1}{k!l!}\sum_{\sigma\in S}\text{sign}(\sigma)\left[\omega(X_{\sigma(1)},\dots,X_{\sigma(k)}), \eta(X_{\sigma(k+1)},\dots,X_{\sigma(k+l)} ) \right]$$
When choosing a basis $\{T_i\}$ for the Lie algebra $\mathfrak{g}$, we have that $\eta$ and $\omega$ can be written as:
$$\eta=\sum_{i=1}^n \eta^i\otimes T_i$$
$$\omega=\sum_{i=1}^n \omega^i\otimes T_i$$
where each $\omega^i$ and $\eta^i$ are $k$ and $l$ forms on $P$ respectively. Then the wedge product can be written as
$$\omega \wedge \eta =\sum_{i,j=1}^n \omega^i\wedge\eta^j\otimes[T_i, T_J] $$
Hence:
$$[\eta,\omega] =\sum_{i,j=1}^n \eta^j\wedge\omega^i\otimes[T_j,T_i] $$
$$=(-1)^{lk}\sum_{i,j=1}^n\eta^i \wedge\omega^j\otimes[T_j,T_i]$$
$$=(-1)^{lk+1}\sum_{i,j=1}^n\eta^i \wedge\omega^j\otimes[T_i,T_j]$$
$$=(-1)^{lk+1}[\omega,\eta]$$
So for two one forms, $\eta$ and $\omega$:
$$[\eta,\omega]=(-1)^2[\omega,\eta]=[\omega,\eta]$$
Hence the wedge product of Lie algebra valued forms is symmetric for one forms.
Best Answer
Here is an easier proof: (Following your notation, it seems that you are using Hamilton's book on mathematical gauge theory, right? I try to stick to this notation too).
First of all, recall that the curvature $F^{A}\in\Omega^{2}(P,\mathfrak{g})$ is "of type $\mathrm{Ad}$'', what essentially means that $$r_{g}^{\ast}F^{A}=\mathrm{Ad}_{g^{-1}}\circ F^{A},$$ where $r_{g}:P\to G$ is the right group action $r_{g}(p):=p\cdot g$. In other words, the curvature is an element of the subset $\Omega^{2}_{\mathrm{hor}}(P,\mathfrak{g})^{\mathrm{Ad}}$.
The secons thing to realize is that
$$F^{f^{\ast}A}=f^{\ast}F^{A},$$
which follows from the fact that $\ast$ commutes with both $\mathrm{d}$ and $[\cdot,\cdot]$. As a next step, note that $f$ can be written as
$$f(p)=p\cdot\sigma_{f}(p)=r_{\sigma_{f}(p)}(p).$$
Hence, combining everything from above, we have that
$$F^{f^{\ast}A}_{p}=(f^{\ast}F^{A})_{p}=(r^{\ast}_{\sigma_{f}(p)}F^{A})_{p}=\mathrm{Ad}_{\sigma_{f}^{-1}(p)}\circ F^{A}_{p}.$$
Last but not least, note that $\sigma_{f}^{-1}=\sigma_{f^{-1}}$ and hence, we get your claimed equality
$$F^{f^{\ast}A}=\mathrm{Ad}_{\sigma_{f^{-1}}}\circ F^{A}.$$