Question: If D4 is acting on the subgroup of its rotations, C4, by conjugation, what are the orbits?

I believe that the orbit of each rotation is itself and its own inverse rotation and nothing else. For the identity, it's just the identity.
Is this correct? What's a good way to show/ conclude this? I determined this through drawing out the actions.

Answer: 
For quickly finding the limit; apply L'Hospital's rule to $$\lim\_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}}$$

To get $$\lim\_{x \to 0} \frac{\ln(x)}{\frac{1}{\sqrt x}} =\lim\_{x \to 0} \frac{\frac 1x}{\frac{-1}{2 x \sqrt x}}=\lim\_{x \to 0} \left(-2\sqrt x \right)=0$$
To get rid of $\sqrt{x}$ you could substitute $u=\sqrt{x}$.

Then $\displaystyle I=\int\_0^1 \sqrt{x}\ln(x)\,dx=\int\_0^1 u\ln(u^2)2u\,du=4\int\_0^1 u^2\ln(u)\,du$

Notice that $u\mapsto u^2\ln(u)$ is continuous on any $[\delta,1]$ with $0<\delta\ll 1$ so the part $\displaystyle \int\_{\delta}^1 u^2\ln(u)\,du$ is finite.

Now in the interval $]0,\delta[$ let's study $f(u)=u\ln(u)$.

$f'(u)=\ln(u)+1\to-\infty$ so $f\searrow$ in a neighborhood of $0^+$ namely on $]0,\delta[$ for a small enough $\delta$.

We deduce that $f$ is bounded on this interval: $|u\ln(u)|<C$

[i.e. *in simple words you cannot decrease from $-\infty$, in math language $g(v)=f(\frac 1v)\nearrow$ and upper bounded $g(v)\le 0$ thus converges when $v\to+\infty$, and a convergent continuous function is bounded.* ].

So $\displaystyle \bigg|\int\_0^\delta u^2\ln(u)\,du\bigg|\le \int\_0^\delta u|u\ln(u)|\,du\le \int\_0^\delta Cu\,du=\frac 12C\delta^2\ll 1$