Isomorphism class of vector bundle over $\mathbb S^1$.

$\require{AMScd}$Consider the map $f:t\in[0,1]\mapsto\exp(2\pi it)\in S^1$. If $E$ is a vector bundle in $S^1$, then $f^*(E)$ is a vector bundle on $[0,1]$. It is not difficult to show that all vector bundles on $[0,1]$ are trivial. You can see from this that there is a continuous bundle map $F$ such that the diagram \begin{CD} [0,1]\times\mathbb R^n @>F>> E \\ @Vp_2VV @V\xi VV \\ [0,1] @>f>> S^1 \end{CD} commutes and $F$ is an isomorphism on fibers. In particular, we can construct a unique linear isomorphism $\phi_f:\mathbb R^n\to\mathbb R^n$ such that $F(1,\phi_f(v))=F(0,v)$ for all $v\in\mathbb R^n$.

Deduce from this that all vector bundles on $S^1$ are obtained from a trivial vector bundle $[0,1]\times\mathbb R^n\to[0,1]$ by identifying $(0,v)$ with $(1,\phi(v))$ for a fixed automorphism $\phi:\mathbb R^n\to\mathbb R^n$.

Next, show that if $\phi$ and $\psi$ are two automorphisms of $\mathbb R^n$ in the same path component of $GL(n,\mathbb R)$, then the correspinding bundles are isomorphic.