Let $B$ be a vector bundle over a compact metric space $X$. Is there always a linear bundle automorphism that covers $f \in \mbox{Homeo}(X)$? Let $B$ be a vector bundle over a compact metric space $X$. Is there always a linear bundle automorphism that covers $f \in \mbox{Homeo}(X)$? 
By covering I mean that $fp=pF$, where $p: B \rightarrow X$ is the bundle projection. 
The collection of linear bundle maps covering $f \in \mbox{Homeo}(X)$ form a Banach space, where the Banach space is determined up to isomorphism by picking a continuous norm on $B$ and defining 
$$|F| = \sup_{x \in X}\sup_{|v| = 1}|F(v)|_{x}$$ 
where $p(v) = x$. As a follow up question, is this Banach space infinite dimensional (if it's non-empty)? 
 A: No. For a counter-example consider $X = S^1\vee S^1$ and $B=\gamma\vee \mathbb{\underline{R}}$ where $\gamma$ is non-trivial, and let $s\colon X\to X$ be the homeomorphism which swaps the two circles. If $s$ is covered by a linear isomorphism then in particular $s$ restricted to the first circle would give an isomorphism between $\gamma$ and $\mathbb{\underline R}$. 
You can slightly rephrase by noticing $f$ is covered by a linear isomorphism iff there is a linear isomorphism $F\colon f^*B\cong B$ over $X$ (i.e. covering the identity map), and there is usually no reason for this to be true for an arbitrary $f$. In particular, if $X$ is a classifying space then $f^*E \cong E$ iff $f$ is homotopic to the identify map. 
Using this perspective you can also do a Stiefel-Whitney class argument for the above example. If $s$ is covered by a linear isomorphism then $w_1(B) = w_1(s^*B) = s^*(w_1(B))$, but 
$$ w_1(B) = (1,0)\in H^1(X;\mathbb{Z}/2)\cong \mathbb{Z}/2 \oplus \mathbb{Z}/2 $$
and $s^*(1,0) = (0,1)$. (This argument is basically equivalent to the first.)

As for the dimension of the Banach space, I'm not sure.
