I think this question is really of the following form: if $f_1,f_2: S^1 \to X$ represent elements $\alpha, \beta$ in $G=\pi_1(X,x_0)$, and $f_1,f_2$ are freely homotopic, then $\alpha, \beta$ are conjugate in the group $G$. Here freely homotopic means not respecting the base point.
So if $a$ is the base point of $S^1$, then $f_1(a)=f_2(a)=x_0$, but the homotopy determines an element $\gamma$ of $G$ represented by the path, say $k$, of the base point during the homotopy. Then $\gamma$ conjugates $\alpha$ to $\beta$ in $G$.
There are several ways to see this. One is to represent the homotopy as a square in which the bottom and top edges are given by $f_1,f_2$ and the vertical edges are given by $k$, since the square identifies to a cylinder to give a homotopy of maps of $S^1$. Because the square is convex, it determines a homotopy which gives the equation $\alpha \gamma= \gamma \beta$.