= Solution
Let $N(t)$ count roots in the open <unit disk>. A boundary crossing can occur only at the real values $0,\pm1,\pm2$ found above, so the count is constant between them. Also $N(-t)=N(t)$, because the <polynomial> is odd. At $t=0$, $p(z)=z(z^4+1)$ has one root inside and four on the <circle>, giving $N(0)=1$.
For small positive $t$, the root near zero stays inside. At each boundary root $\zeta^4=-1$, $p'(\zeta)=-4$, so implicit differentiation gives $z(t)=\zeta-t/4+O(t^2)$. The <radial crossing test for polynomial root counts> is
$$
\left.\frac d{dt}\log|z(t)|\right|_{t=0}=-\frac{\operatorname{Re}\zeta}{4}.
$$
The two roots with positive real part enter and the other two leave. Hence $N(t)=3$ for $0<t<1$.
At $t=1$, the boundary roots are $\zeta=e^{\pm i\pi/3}$. Since $\zeta p'(\zeta)=5t-4\zeta$, their radial derivative is $\operatorname{Re}(1/(5-4\zeta))=1/7>0$. Both leave as $t$ increases, so the count becomes one; at $t=1$ itself they are excluded, also leaving one interior root. At $t=2$, the remaining boundary root $z=1$ has radial derivative $1/6>0$ and leaves. For $t\ge2$ no interior root is possible, since $|p(z)|\le |z|^5+|z|<2$ in the open <disk>. Combining these facts and odd symmetry gives
$$
\boxed{N(t)=\begin{cases}
3,&0<|t|<1,\\
1,&t=0\ \text{or}\ 1\le|t|<2,\\
0,&|t|\ge2.
\end{cases}}
$$
This also gives the corresponding <winding numbers> of the <image> <contour> away from its crossings. Boundary roots are excluded at the crossing values; using a nearby <winding number> at $t=0$ would incorrectly count two extra roots.
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