Are vector bundles isomorphic when their transition functions are homotopic?

I'm in the process of trying to understand the equivalence between two different approaches to vector bundles, namely defining them to be smooth surjections $E\to B$ that are locally isomorphic to projections $U_\alpha\times\mathbb{R}^n\to U_\alpha$, and defining them to be collections of transition functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\to\mathrm{GL}_n(\mathbb{R})$ which satisfy the trivial ($g_{\alpha\alpha}=\mathrm{id}$) and cocycle ($g_{\beta\gamma}g_{\alpha\beta}=g_{\alpha\gamma}$) identities.

Where I'm stuck right now is in trying to prove the following, intuitively true, statement: given an open cover $B=\bigcup U_\alpha$ and two collections of transition functions $g_{\alpha\beta},h_{\alpha\beta}:U_\alpha\cap U_\beta\to\mathrm{GL}_n(\mathbb{R})$ for which the $g_{\alpha\beta}$ are homotopic to the $h_{\alpha\beta}$ via a family of transition functions, the vector bundles they define are isomorphic. To rephrase my hypotheses, I'm saying that we have homotopies $g_{\alpha\beta}^t:(U_\alpha\cap U_\beta)\times I \to \mathrm{GL}_n(\mathbb{R})$ from $g_{\alpha\beta}^0=g_{\alpha\beta}$ to $g_{\alpha\beta}^1=h_{\alpha\beta}$, and for each $t\in I$, the $g_{\alpha\beta}^t$ form a family of transition functions.

I find this to be intuitively true because "vector bundles being isomorphic" to me is a "rigid" property, so that if $E_t$ is the vector bundle defined by the $g_{\alpha\beta}^t$, then if $t=t_0$ were the first point at which the isomorphism class of $E_t$ changes, we should expect some problem to arise there.

So, I'd like to ask for a proof of my claim if it's true, or a counterexample if it is false. If I could ask further, what about the converse? Also, what happens if we remove the requirement that the $g_{\alpha\beta}$ be homotopic to the $h_{\alpha\beta}$ via a family of transition functions, and just require that for each $\alpha,\beta$, they are homotopic?

Given a «homotopy of transition functions» $(g_{\alpha,\beta}^t)$ you can construct a vector bundle on $B\times [0,1]$ whose restriction to $B\times\{0\}$ and to $B\times\{1\}$ are your vector bundles $E_0$ and $E_1$.