The question is updated as following.
1. Let $(\Phi,L^2(R))$ be left regular representation of $\mathbb R$ given by $$ \Phi(g)f(x)=f(x-g). $$ It is unitary representation of $\mathbb R$.
2. For $G=\mathbb R$, $c_s(t)=ts,s\in\mathbb R$ is $1$-parameter subgroups in $\mathbb R$. Note $c_s'(0)=s$. Thus $Lie(\mathbb R)=\{s\in\mathbb R\}$ which is $1$-dimensional vector space over $\mathbb R$.
3. For $f\in C_c^\infty(\mathbb R)$, consider the action of $X\in Lie(\mathbb R)$, $$ \varphi(X)f(x)=\frac{d}{d t}\Phi(t X)f(x)_{t=0}=\frac{d}{d t}f(x-tX)_{t=0}=f'(x)(-X). $$ Note that $\Phi$ is unitary. The action of $\varphi(X)$ is skew-Hermitian, which implies that $\varphi(X)f(x)$ should be a imaginary multiples of $f'(x)$.
What's wrong?