Let $\xi=(E,p,B),\xi'=(E',p',B')$ be fibre bundles. Let $f: B\to B'$, $\bar f: E\to E'$ be maps such that the diagram commutes
$\require{AMScd}$
\begin{CD}
 E @>\displaystyle\bar f>> E'\\
 @V \displaystyle p V V\# @VV\displaystyle p' V\\
 B @>>\displaystyle f> B^{\,\prime}
 \end{CD}

Does this alway imply that $\xi$ is isomorphic to the pull-back bundle ( the pull-back bundle, denoted by $\eta$, of $\xi'$ to $B$ through $f$ is: $E(\eta)=\{(b,e')\mid b\in B, e'\in E' \text{ such that } f(b)=p'(e')\}$, $p(\eta)(b,e')=b$, $B(\eta)=B$) of $\xi'$?