Let $G$ be a finite group. $G$ can bear the so-called regular representation. Let $\chi_g(h) \colon= \delta_{g,h} ~ {\mathrm{for}} ~ h \not= g$. Let $X \colon= {\mathrm{the\, vector\,space\,of\,dimension}}\,|G|\,{\mathrm{over}}\,{\Bbb C} ~{\mathrm{spanned~by}} ~\chi_{g}{\mathrm{s}}$. There are two actions on $X$.
One is defined by $gx(h) \colon = h(g^{-1}h)$ and another $gx(h) \colon= x(hg)$.
Q. Suppose we fix the base of $X$ and realise these two actions by matrices. Then are these two representations conjugate? That is, are they similar up to some inner automorphism?
Note that your space $X$ can easily be identified with the set of $\mathbb{C}$-valued functions on $G$ (and it seems to me you have already done so yourself).
Let us write $(g.x)(h) = x(g^{-1}h)$ and $(g \cdot x)(h) = x(hg)$ to distinguish the two actions.
Consider the map $S: X \to X$ mapping $x: G \to \mathbb{C}$ to $S(x): G \to \mathbb{C}$ defined by $S(x)(g) = x(g^{-1})$. Note that $S$ is linear and $S^2 = id$, so $S$ is bijective.
Now note that $$(g \cdot S(x))(h) = S(x)(hg) = x(g^{-1} h^{-1}) = (g.x)(h^{-1}) = S(g.x)(h)$$ for all $g,h \in G$ and $x \in X$. It follows that $S(g.x) = g \cdot S(x)$ and so the representations are indeed equivalent.