Let $G$ be a topological group, let $f:X\rightarrow Z$ be a $G$-equivariant map of (left) $G$-spaces such that 


1. $X\rightarrow X/G$ and $Z\rightarrow Z/G$ are principal $G$-bundles.
2. $f$ is a fibration.


Let $\rho: G\rightarrow H$ be a morphism of topological groups. Is the induced map 
$$ H\times\_{G}X \rightarrow H\times\_{G}Z$$ a fibration ? where the (right) action of $G$ on $H$ is induced by $\rho$


**Edit:** I was trying the following proof, but I think it is incomplete may be someone could help. 


First: $f$ induces a continuous map of topological spaces $\hat{f}: X/G\rightarrow Z/G$.
The pullback of the map $Z\rightarrow Z/G$ along $\hat{f}: X/G\rightarrow Z/G$ is exactly (up to isomorphism) the map $X\rightarrow X/G$. 


Second: The pullback of the map $H\times\_{G}Z\rightarrow Z/G$ along $\hat{f}$ is exactly (up to isomorphism) the map $H\times\_{G}X\rightarrow X/G$.


In the first and second item we use the fact that we have $G$-principal bundles and $H$-principal bundles. 


If we can proof that $\hat{f}$ is a fibration then we are done! So my question would be answered if $\hat{f}$ is a fibration. If $f$ is a fibration does it follow that $\hat{f}$ is a fibration ? 



Fibration of principal bundles
The previous answer was getting a bit too complicated. Locally $f$ looks like $O\_i \times G \rightarrow U\_i \times G$ and this is a fibration since it is the restriction of the original fibration $f$. Now $\hat f$ locally looks like $O\_i \rightarrow U\_i$, and this is a fibration since it is a retract of a fibration. Thus locally $\hat f$ is a fibration, and so, thanks to Dold, $\hat f$ is a fibration.


