Consider $f,\overline{f}:\left(S^1, s_0\right) \to (X,x_0)$ two paths in $X$ such that $\overline{f}=f \circ r$, where $r: S^1 \to S^1$ is the reflection around the $1$-space with $x_2 = 0$.
I would like to prove that
$$f*\overline{f}\big(r(x)\big)=f*\overline{f}(x)$$
Where "$*$" is the concatenation of paths.
Let $S^1\subseteq\mathbb{C}$ be the unit circle. Then the reflection map $r:\mathbb{C}\rightarrow\mathbb{C}$ is just complex conjugation $z\mapsto \overline z$. Concatenation of two based loops $f,g:S^1\rightarrow X$ is defined by $$(f\ast g)(z)=\begin{cases}f(z^2)& Im(z)\geq0\\ g(z^2)&im(z)\leq0.\end{cases}$$ For a based loop $f:S^1\rightarrow X$ the map $\overline f:S^1\rightarrow X$, $z\mapsto f(\overline z)$ is also a based loop and we have $$(f\ast \overline f)(\overline z)=\begin{cases}f(\overline{z}^2)& Im(\overline z)\geq0\\ \overline f(\overline z^2)&im(z)\leq0\end{cases}=\begin{cases}f(\overline{z}^2)& Im( z)\leq0\\ f(z^2)&im(z)\geq0\end{cases}$$ which is equal to $(f\ast\overline f)(z)$.