Left divisor of a fibration by compact Lie group is a fibration.

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Let $p:E \rightarrow B$ be a Hurewicz fibration where E and B are path-connected and compact CW complexes. Let $G$ be a compact Lie group (left) acting on $E$ and let $E’$ be the resulting orbit space of that action, i.e., $E’=E/G$. Suppose we have an induced map $f:E’ \to B$ such that $p = fq$, where $q:E \to E’$ is the canonical quotient map. Can we say that $f:E’ \to B$ is a Hurewicz fibration? Can we at least expect $f:E’ \to B$ to be a Serre fibration, i.e., have homotopy lifting property with respect to $[0,1]^n$ for all $n \in \mathbb{N}$? I know that this is true if the action of $G$ on $E$ is free, but my action here is NOT free.

To prove this, all we need to show is that any given map $\phi’_{n}:[0,1]^n \to E’$ can be “lifted” to a map $\phi _{n}:[0,1]^n \to E$ for each $n$ so that $q \phi _{n} = \phi’ _{n} $. I know this can be done when $n=1$ (see Bredon’s Introduction to compact transformation groups: Lemma 6.1 and Theorem 6.2 on Pages 90-91). I suspect that Bredon’s Lemma 6.1 (and hence Theorem. 6.2) can be extended to all cubes, i.e., to $[0,1]^n$ for all $n$, however, I am not sure.

Any help or references towards proving that $f:E’ \to B$ is a Serre fibration will be appreciated.