Let $G$ be a topological group, and let $F$ be the Markov free topological group over $G$. We define an action of $G$ on $F$ as follows: $G\times F\rightarrow F$, $^{g}(g_{1}^{\epsilon_{1}}...g_{n}^{\epsilon_{n}})$=$(^{g}g_{1})^{\epsilon_{1}}...(^{g}g_{n})^{\epsilon_{n}}$. Question: Is the abovedefined action, a contiuous action? Why? (Note that, here $G$ acts on itself by conjugation.)

I don't know the answer to the general question but the answer seems to be "yes" when $G$ is compact. Here is a straightforward argument. The corresponding action of $G$ on the free topological monoid $M=\coprod_{n\geq 0}(G\sqcup G^{1})^n$ is certainly continuous. When $G$ is compact $F(G)$ is the topological quotient of $M$ via word reduction $r:M\to F(G)$. When the action of $G$ on itself is conjugation, we have $^{g}(r(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))=r(^{g}(g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}))$ for any $g_{1}^{\epsilon_1}...g_{n}^{\epsilon_n}\in M$. The action is continuous since $id\times r:G\times M\to G\times F(G)$ is a topological quotient map (compactness of $G$ helps out here too). This argument should also work when $G$ is a $k_{\omega}$space (inductive limit of nested compact subspaces) but doesn't extend to the general case since $r:M\to F(G)$ is not quotient when $G=\mathbb{Q}$ is the rationals. 

