This is Exercise 36 of section 4.3 (p. 132) from Abstract Algebra by Dummit & Foote.
Let $\DeclareMathOperator{\Sym}{Sym} \pi_1:G\rightarrow \Sym(G)$ be the left regular representation afforded by the action of $G$ on itself by left multiplication defined by $\pi_1(g)=\sigma_g$ so that $\sigma_g(x)=gx$ for all $x\in G$.
Let $\pi_2:G\rightarrow \Sym(G)$ be the right regular representation afforded by the action of $G$ on itself by right multiplication defined by $\pi_2(h)=\tau_h$ so that $\tau_h(x)=xh^{-1}$ for all $x\in G$.
Prove that $\sigma_g=\tau_h$ iff $g,h\in Z(G)$. Deduce that $\pi_1(G)\cap\pi_2(G)=\pi_1(Z(G))=\pi_2(Z(G))$
The 'only if ' part is as follows: Suppose $\sigma_g=\tau_h$. For all $x\in G$, $\sigma_g(x)=gx=xh^{-1}=\tau_h(x)$. In particular, $gh=hh^{-1}=1$, so $h=g^{-1}$. So $\sigma_g(x)=gx=xg=x(g^{-1})^{-1}=\tau_{g^{-1}}(x)$ for all $x\in G$. So $g\in Z(G)$. Also $h=g^{-1}\in Z(G)$.
As you can see, the 'only if' part is quite easy. What's frustrating me is the 'if' part. Whenever $g$ and $h$ lie in the center of $G$, why is $\sigma_g=\tau_h$? This is so counter-intuitive: Since $1\in Z(G)$, by letting $g\in Z(G)$, $g\neq 1$, we get $gx=x$ for all $x\in G$, and so $g=1$! And so $G$ is centerless!