Given trivial fiber bundles $\pi : B \times F \rightarrow B$ and $\pi' : B \times F' \to B$ and an isomorphism $h : B \times F \to B \times F'$ of fiber bundles (i.e. so that $\pi = \pi'h$), we know that for a fixed $b$ in $B$, the map $h$ restricted to $\{b\} \times F$ can be corestricted to $\{b\} \times F'$ and so, in particular, we have a homeomorphism $$F \simeq \{b\} \times F \simeq \{b\} \times F' \simeq F', $$ given by this restriction and the restriction of $h^{-1}$ to $ \{b\} \times F'$. Moreover, we see that $h(b,f) = (b,g(b,f))$ with $g(b,-)$ a homeomorphism for each fixed $b \in B$.
So, my question is, should $\pi_2h$ depend only on the second variable, so that we have a function $g'(f) = g(b,f)$ for all $b$ and thus $h(b,f) = (b,g'(f))$ with $g : F \to F'$ a homeomorphism?
Your question is whether each isomorphism $h : B \times F \to B \times F'$ of trivial fiber bundles has the form $h = id_B \times h'$ with a homeomorphism $h' : F \to F'$.
The answer is no. Consider for example $B = [0,1]$ and $F = F' = \mathbb{R}$. Define $h(t,x) = (t,(1+t)x)$. This is an isomorphism of trivial vector bundles which has not the desired form.
The general background is that (under mild assumptions on the spaces involved) we can identify bundle isomorphisms $h$ with maps $h^* : B \to Iso(F,F')$, where $Iso(F,F')$ is the "space of isomorphisms between the fibers". We shall not make this precise; what $\mathcal{I} = Iso(F,F')$ is depends on the special context, but in general this space admits non-constant maps $B \to \mathcal{I}$.
In the above example we have $Iso(\mathbb{R},\mathbb{R}) = GL(\mathbb{R},1)$.
Note that even if the fibres are as simple as $\mathbb{Z}_2$ (with the discrete topology), we get non-constant maps $B \to \mathcal{I}$ if $B$ is not connected.