Let $G$ be a topological group and $M,N$ be $G-spaces$ (CW-complexes). G acts freely on $M$. Then
$$(N \times M)/G \to M/G$$
is a fibration with all fibers homeomorphic to $N$.
Thoughts: Of course $$N \times M \to M$$ is a fibration with all fibers homeomorphic to N. Is that enough to conclude that our map is a fibration at least? Let's take $[m] \in M/G$. Its fiber is $[(n,g\cdot m)]$. We can of course remove the $g$ because $[(n,g \cdot m)] = [(g^{-1}n,m)]$. Now $$ [(n_1,m)]=[(n_2,m)] \implies g\cdot(n_1,m)=(n_2, m)$$ hence $g\cdot m = m$ but $G$ acts freely on $M$ so $g=1$ and $n_1=n_2$ so the fiber is homeomorphic to $N$.
Consider the commutative diagram of quotient maps
Now $q^{-1}([m])=N\times Gm$. Since $q=rp$ and $p$ is surjective, we have $$ r^{-1}([m]) = p(N\times Gm) $$
Because $p(n,gm)=p(g^{-1}n,m)$, we see also that $$ r^{-1}([m]) = p(N\times\{m\})$$
Consider the map $s: N\rightarrow p(N\times\{m\})$, given by $s(n)=p(n,m)$. This is surjective and continuous; you've already shown it is injective because $G$ acts on $M$ freely. If $G$ acts properly discontinuously, let $V$ be an open set of $M$ such that $V\cap Gm=\{m\}$. Then if $U\subset N$ is open, we have $$ s(U) = p(U\times\{m\})=p(N\times\{m\})\cap p(U\times V)$$ which is open in $p(N\times\{m\})$ because $p$ is an open map. Thus $s$ is a homeomorphism.
Edit: we don't need to assume $G$ acts properly discontinuously on $M$. We already have a continuous bijection $s:N\rightarrow p(N\times\{m\})$. The set-theoretic inverse $s^{-1}$ is given by $$ s^{-1}([n,gm]) = g^{-1}n $$
For this to be continuous, for any point and neighborhood $n\in U\subset N$, we need neighborhoods $n\in W\subset N$ and $m\in V\subset Gm$ such that $$ \bigcup_{gm\in V} gW\subset U $$
When $G$ is acting properly discontinuously on $M$, we can take $W=U$ and $V=\{m\}$. When $G$ acts trivially on $N$, we can take $W=U$ and $V=Gm$. More generally, we can always find $W$ and $V$ from the joint continuity of the action $G\times N\rightarrow N$, provided $G\rightarrow Gm$ is an embedding. This is true when $G$ is compact, or if $G$ is a Lie group acting properly on $M$. But it's not always true: for example, if $G$ is $\mathbb{R}$ and $M$ is a torus, with $Gm$ a dense irrational curve.