Let $G$ be a compact topological group and $f:G\to G$ a continuous homomorphism. Assume there is a dense subset $X\subset G$, such that that for all $x\in X$, we have that $f(x)$ is conjugate to $x$. Is it then true that $f(g)$ is conjugate to $g$ for all $g\in G$?
If $G$ were a metric space, we could make a sequence argument. Pick some sequence $(x_n)_n \in X$ converging to $g$. Then there are elements $(g_n)_n \in G$ such that
$$f(x_n)=g_nx_ng_n^{-1}. $$ Because $G$ is compact, we can pick a converging subsequence of $(g_n)_n$ with limit $g_0$. Then $$f(g)=g_0gg_0^{-1}.$$
I am not sure how to adapt this argument to the case where $G$ is not a metric space.
As Moishe Kohan suggested, one may use nets, if we add the assumption that $G$ is Hausdorff. Fix $g\in G$. For each open $U\subset G$ containing $g$, we may pick am element $x_U \in X \cap U$. We obtain two nets $(x_U)_U \subset X$ and $(g_U)_U \subset G$, such that $$f(x_U)=g_Ux_Ug_U^{-1}.$$ Using the compactness assumption, we may pass to subsequences and assume that both nets converge. Clearly $(x_U)_U$ converges to $g$. Because $f$ is continuous $(f(x_U))_U$, converges to $f(g)$. Let $g_0$ be the limit of $(g_U)_U$.
$G$ being Hausdorff, implies that limits of nets are unique. If we prove that $(f(x_U))_U$ converges to $g_0gg_0^{-1}$, we may conclude that $f(g)=g_0gg_0^{-1}$.
The function $$\psi:G^3\longrightarrow G,\psi(g_1,g_2,g_3)=g_1g_2g_3^{-1}$$ is continuous and the limit of the net $(g_U,x_U,g_U)_U \subset G^3$ is $(g_0,g,g_0) \in G^3$. Therefore the limit of $(f(x_U)_U)$ is $g_0gg_0^{-1}$.