Note: this question is really a subquestion of this one, but I decided to ask it separately since it seems it might be attacked first.
Let $B$ be a topological space and $G$ a topological group acting on a space $F$. Then the space $G^B$ of continuous maps from $B$ to $G$ has a group structure given by pointwise multiplication. Moreover, for nice enough spaces the compact-open topology gives $G^B$ the structure of a topological group. This group acts on the space $F^B$ (again endowed with the compact open-topology) in the obvious pointwise way and the action is continuous.
My question is the following: assume $G$ acts transitively; can we infer the same holds for $G^B$? If not, under what assumptions is transitivity preserved?
Notice that freeness and faithfulness are (separately) preserved (it is enough to think pointwise).
No. Consider the case $G=(\Bbb R,+),~B=F=S^1$, and $G$ acts on $F$ via $$t\cdot z=\exp(it)\cdot z$$ Then there is no map $g:B\to G$ such that $g\cdot 1=\mathrm{id}$, where $1$ is the constant map $B\to F, z\mapsto 1$. Indeed any map $g: S^1\to\Bbb R$ is homotopic to a constant map because $\Bbb R$ is contractible, however $1$ and $\mathrm{id}$ are not homotopic as maps $S^1\to S^1$.
More precisely, in this example we have that two maps $B\to F$ are in the same $G^B$ orbit iff they are homotopic as maps $S^1\to S^1$.