One of the fundamental facts in fiber bundle theory is the following result for existence and extension of sections (see Thm. 9 in this paper of Palais, and compare with Thm. 12.2 in Steenrod's book):

**Theorem:** *Let $\pi\colon E\to B$ be a locally trivial fiber bundle with fiber $F$ a contractible metrizable manifold, and base space $B$ a metrizable space. Let $A$ be a closed subspace of $B$ and $\sigma\colon A \to E$ be a continuous section of $E$ over $A$. There is then a continuous extension of $\sigma$ to a global section of $E$.*

I would like to know if the above result has some equivariant version, for instance something along the following lines:

Question:Let $\pi\colon E\to B$ be a locally trivial fiber bundle with fiber $F$ a contractible metrizable manifold, and base space $B$ a metrizable $G$-space. Suppose also that $E$ is a $G$-space and $\pi$ is equivariant. Let $A$ be a closed $G$-invariant subspace of $B$ and $\sigma\colon A \to E$ be a continuous $G$-equivariant section of $E$ over $A$. Does there exist a continuous extension of $\sigma$ to a global $G$-equivariant section of $E$?

If the answer in general is no, then what are some additional conditions which would ensure the existence of an equivariant section?

**Addendum:** If the global section $\sigma$ exists, then for each $b\in B$, we must have
$
\sigma(b)\in F_b^{G_{F_b}},
$
where $F_b:=\pi^{-1}(b)$ is the fiber over $b$, $G_{F_b}$ is the subgroup of $G$ mapping $F_b$ to itself, and $F_b^{G_{F_b}}$ is the set of points in $F_b$ fixed by $G_{F_b}$. So a necessary condition for the existence of $\sigma$ would be
$$
F_b^{G_{F_b}}\neq\emptyset
$$
for all $b\in B$. Further, it would be reasonable to assume that $F_b^{G_{F_b}}$ is contractible in parallel to the above theorem; or, to simplify things even further, assume that $F^H$ is contractible for all subgroups $H$ of $G$. Would this condition be enough to ensure the existence of $\sigma$?