Is a bundle morphism which restricts to homeomorphisms of the fibers a bundle isomorphism? If $f$ is such a map between total spaces (assume a common base space) then it is a bijective and continuous and the inverse will be fiber preserving so that all we would need to prove is that the inverse is continuous. This question came to me because of an easy lemma of Hatcher proves in "Vector bundles" which states
"A continuous map $f : E \rightarrow E'$ between vector bundles over the same base space B is an isomorphism if it takes each fiber $p_1^{−1}(b)$ to the corresponding fiber $p^{−1}_2(b)$ by a linear isomorphism"
Hatcher proves this by composing with trivialisations so that $f^{-1}$ being continuous is equivalent to the contiuity of $(x,g^{-1}(x)(v))$ where $(x,g(x)(v))$ is the map between trivialisations induced by $f$ (i.e. $h'\circ f \circ h^{-1}(x,v)=(x,g(x)(v))$.
Asking the question about a general fiber bundle and attempting the proof in the same way leads to the question
"Let $\lambda(x)$ be a collection of homeomorphisms of $F$ to itself which are contiuous in $x$ in the sense that $(x,\lambda(x)v)$ is continuous from $X \times F$ to $X \times F$. Then is the collection $(x,\lambda(x)^{-1}v)$ contiuous from $X \times F$ to $X \times F$"? Is this true?
 A: Suppose the image of $\lambda$ lies in some group of homeomorphisms $G$, topologized as a subset of the mapping space $F^F$, such that with this topology $G$ is a topological group. Suppose further that the fiber is locally compact. Then there is a continuous evaluation map $F^F \times F \to F$, and because $F^F$ is an exponential object,  your claim that $X \times F \to X \times F$ is continuous is the same as that the map $\lambda: X \to G$ is continuous; then because $G$ is a topological group, $\lambda^{-1}$ is continuous, and again because $F^F$ is an exponential object you have the continuity of the inverse as desired.
When do these conditions hold? If $F$ is also locally connected, in addition to being locally compact and Hausdorff, then $\text{Homeo}(F)$ is a topological group; an improvement of this is given here. In general if $F$ is compact Hausdorff $\text{Homeo}(F)$ is a group. 
So in particular this is true if $F$ is a (separable) finite-dimensional manifold. But it is not true in general: if $\text{Homeo}(F)$ is not a group, then take $X = \text{Homeo}(F)$ and $\lambda = \text{id}$. The second paper linked above gives a proof that inversion is not continuous for $F$ the cantor set minus a point.
