Let $G$ be a topological group, let $f:X\rightarrow Z$ be a $G$-equivariant map of (left) $G$-spaces such that
$X\rightarrow X/G$ and $Z\rightarrow Z/G$ are principal $G$-bundles.
$f$ 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 ?