$\newcommand{\proj}{\operatorname{proj}}$ $\dim_H(\cdot)$ is the Hausdorff dimension of a set. Let us denote the line through the origin in $\mathbb{R}^2$ which makes an angle of $\theta$ with the horizontal by $L_{\theta}$. Moreover, let us define the orthogonal projection onto $L_{\theta}$ by $\proj_{\theta}V$.
QUESTION: Let $E$ be the $`$circular Cantor set$'$ given in complex number form by $E = \{ e^{2\pi i\varphi} : \varphi \in F \}$, where $F$ is the middle third Cantor set. What is $\dim_H\proj_\theta E$ for each $\theta$?
For the first implication I get that:
Let us define the function $f:[0,1] \rightarrow C$, where C is the unit circle in the complex plane by $\begin{equation} f(\varphi) = e^{2\pi i \varphi} \ \ \ \text{where $\varphi \in [0,1]$} \end{equation}$
Now we can show that the function $f$ is Lipschitz mapping as if $\varphi_1$, $\varphi_2 \in [0,1]$ then
$$\begin{equation} |f(\varphi_1) - f(\varphi_2)| = |e^{2 \pi i \varphi_1}-e^{2 \pi i \varphi_2}| \leq 2\pi |\varphi_1 - \varphi_2|. \end{equation}$$
Therefore we obtain that $\begin{equation} \dim_HE = \dim_H(f(F)) \leq \dim_HF = \frac{\log 2}{\log 3} \end{equation}$
So as $\dim_{H}V \leq \text{min} \{\dim_H V,1\}$ for some set $V$, we have that $$ \dim_H(\proj_\theta E) \leq \frac{\log 2}{\log 3} $$
However I cannot see how to do the opposite implication.
The answer is boring. For every subset $E$ of the circle, and for every $\theta$, $$\dim_H\operatorname{proj}_\theta E=\dim_H E\tag{1}$$ You already have the $\le $ inequality. To prove $\ge$, proceed as follows:
By rotating $E$, it suffices to deal with $\theta=0$. Except for the points $\pm 1$, the set $E$ can be covered by countably many closed arcs $\Gamma_n$ not containing $\pm 1$. (Take the connected components of the intersection of the circle with the set $|x|<1-1/n$.) The restriction of the projection map onto each $\Gamma_n$ is not only Lipschitz, but has a Lipschitz inverse (i.e., it is bi-Lipschitz). Therefore, $$\dim_H(\operatorname{proj}_\theta (E\cap \Gamma_n)) = \dim_H(E\cap \Gamma_n) \tag{2}$$
Since the Hausdorff dimension is countably stable, $\dim_H E = \sup_n \dim_H (E\cap \Gamma_n)$. The conclusion follows.