= Solution
Polar deformation reduces maps from the circle into $GL_N$ to maps into $U(N)$. Define their integer invariant by the <winding number> of the determinant. It is unchanged under <homotopy> and stabilization, and determinants multiply under block sum, so it gives a homomorphism $K^1(S^1)\to\mathbb Z$. Every integer occurs, via $z\mapsto\operatorname{diag}(z^m,1,\ldots,1)$.
To prove injectivity, first consider a map with determinant winding zero. Its determinant has a continuous logarithm $ih(z)$ on the circle. Multiplying the first row by $e^{-ith(z)}$ gives a <homotopy> into $SU(N)$ at $t=1$. If $N=1$ that endpoint is already the identity. For $N\geq2$, $SU(N)$ is simply connected: $SU(2)\cong S^3$, and the last-column fibration
$$
SU(N-1)\longrightarrow SU(N)\longrightarrow S^{2N-1}
$$
with locally trivializations from orthonormal completion, together with its <homotopy> exact sequence, proves this inductively because the sphere has zero first and second <homotopy> groups. Thus the remaining circle map is null-homotopic. For a map of winding $m$, multiply by $\operatorname{diag}(z^{-m},1,\ldots,1)$ and apply the zero-winding argument. This proves
$$
\boxed{K^1(S^1)\cong\mathbb Z.}
$$
For higher spheres, stable unitary <homotopy> gives $K^1(S^n)=\pi_n(U)$: a map can be normalized by multiplying by the inverse of its value at the basepoint, a free <homotopy> because the <unitary group> is path connected. A free <homotopy> between based maps can be made based by replacing $H(x,t)$ with $H(x_0,t)^{-1}H(x,t)$, without altering its endpoints. The <Bott periodicity theorem>, in its unitary form $\pi_{j+2}(U)\cong\pi_j(U)$ with $\pi_0(U)=0$ and $\pi_1(U)=\mathbb Z$, gives
$$
\boxed{K^1(S^n)\cong\begin{cases}\mathbb Z,&n\text{ odd},\\0,&n\text{ even}.\end{cases}}
$$
The circle case was proved directly above; the higher-sphere values use the stated general periodicity theorem.
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