= Solution
Let $a,b$ be the two oriented circle generators of the <fundamental group of a bouquet of circles>, $\pi_1(S^1\vee S^1)\cong F(a,b)$. Take
$$
\boxed{[\alpha]=a,\qquad[\beta]=b^{-1}ab}.
$$
Their loops are freely homotopic: slide the basepoint of the $a$ loop along the path $b$, thereby introducing the route $b^{-1}$ to the original loop and $b$ back. Equivalently the change-of-basepoint construction along a loop realizes conjugation, and the retraced connecting portions can be contracted with a moving marked point.
They are not based homotopic, since their <reduced words> in the <free group> are distinct: $a$ has length one and $b^{-1}ab$ has length three with no cancellation. This illustrates that free <homotopy> remembers conjugacy classes, whereas <based homotopy> remembers actual elements of the <fundamental group>.
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