Involution in $L^{1}(G)$ is isometric. (Sorry for asking so many questions of the same type.  There is an underlying issue that I think once resolved will allow me to understand them all at once.)
Let $G$ be a locally compact group, and $f\in L^{1}(G)$.
I'm trying to verify that $\|f^{*}\|_{p} = \|f\|_{p}$, where the involution $*$ is defined by $f^{*}(x) = \delta(x)^{-1}\overline{f(x^{-1})}$.
So my computations so far look like:
\begin{eqnarray*}
\int_{G}f^{*}(y)\mu(dy) &=& \int_{G}\left|\delta(y^{-1})\overline{f(y^{-1})}\right|^{p}\mu(dy)\\
                        &=& \int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy)\\
\end{eqnarray*}
Here I am stuck at the central issue I am facing where the calculations stop making sense and I feel like I'm just throwing things around hoping it's true.
How do I use the definition of the $\delta$ map:
$$\text{the unique group homomorphism }\delta:G\to(0,\infty)\text{ such that }\\\delta(x)\mu(E) = \mu(Ex)\text{ for all Borel sets }E$$
when its argument is not independant of the variable of integration?
Ted has already confirmed in his answer to:
How to use the modular function of a locally compact group?
that the definition of $\delta$ implies:
$$\int_{G}\delta(x)f(y)\mu(dy) = \int_{G}f(y)\mu(dy\cdot x)$$
If I try to use that step in my current problem, I end up with
\begin{eqnarray*}
\int_{G}\delta(y^{-1})^{p}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy) &=& \int_{G}\left|\overline{f(y^{-1})}\right|^{p}\mu(dy\cdot (y^{-1})^{p})
\end{eqnarray*}
which I don't have any clue how to interpret, and I suspect is wrong.  What is the correct way to handle this function?  I dont know what measure to associate with the notation $\mu(dy\cdot (y^{-1})^{p})$.
I think the answer to this question is what I really need to be able to finish the verifications I'm working on.
Thanks to whoever can help me!
 A: I think you would find the last chapter of Real Analysis by Folland helpful (it covers the fundamentals of Haar measures).  The way I think of it, $\delta^{-1}$ is a factor attached to $\mu$ (a left Haar measure), not to $f$. To begin with, the measure $\delta^{-1}\mu$ is right-invariant: 
$$\int_{E\,y}\delta(x)^{-1}\,d\mu(x)=\int_{E}\delta(xy)^{-1}\,\delta(y)\,d\mu(x)= \int_{E}\delta(x)^{-1} \,d\mu(x) $$
And here is another way to get a right invariant measure from $\mu$. 
Let $\iota:G\to G$ be the inversion, $\iota(x)=x^{-1}$. This is an anti-homomorphism, it changes the order of group multiplication. Therefore,  the pushforward $\iota^*\mu$ is a right Haar measure. 
Fact: $\delta^{-1}\mu =  \iota^*\mu$. 
Proof. They are both right Haar measures, therefore one is a constant multiple of each other. To see that the multiple must be $1$, consider a symmetric neighborhood of identity, called $U$, which is small enough so that $|\delta(x)^{-1}-1|<\epsilon$ for $x\in U$. (Such $U$ exists because the modular function is continuous.) Now, $\iota^*\mu(U)=\mu(U)$ by symmetry, and also $|\delta^{-1}\mu(U)-\mu(U)|<\epsilon \,\mu(U) $. Since the ratio of $\iota^*\mu(U)$ to $\delta^{-1}\mu(U)$ can be made arbitrarily close to $1$, but is also a constant (independent ot $U$), the ratio is in fact $1$. $\Box$
The rest is a tautology from abstract measure theory: $$\int (f\circ \iota) \,d\mu = \int f \,d (\iota^*\mu) \tag1$$
hence $$\int f(x^{-1})\,d\mu(x) = \int f(x)\,\delta(x)^{-1} \,d\mu(x) \tag2$$
Actually,   you want to put   $f\circ \iota$ instead of $f$ here, so that the result is exactly as required:
$$\int f(x )\,d\mu(x) = \int f(x^{-1})\,\delta(x)^{-1} \,d\mu(x) \tag3$$

But I don't think your construction produces an isometry on $L^p$, at least not as stated. The identity (3) immediately generalizes to 
$$\int |f(x )|^p\,d\mu(x) = \int |f(x^{-1})|^{p}\delta(x)^{-1} \,d\mu(x) \tag4$$ but if you rewrite (4) by attaching $\delta^{-1}$ to $f$, it has to be raised to power $1/p$. 
