I am aware that uniqueness of a Haar measure on a locally compact topological group is a well known fact in the sense below:
"If $v,\mu$ are Radon measures that are left invariant then there exists a positive constant $a$ such that $v=a\mu$. (Where a Radon measure is a locally finite Borel measure that satisfies the weak inner regularity and outer regularity conditions)."
For the Lebesgue measure on $\mathbb{R}^n$ the above can be strengthened in the sense that the assumptions of regularity are redundant. So, the following holds:
"If $\mu$ is a locally finite Borel measure on $\mathbb{R}^n$ such that $\mu(A+x)=\mu(A)$ for any $A \in \mathcal{B}\left(\mathbb{R^n}\right)$ and $x \in \mathbb{R^n}$ then $\mu=a\lambda$, where $a\geq0$ and $\lambda$ is the Lebesgue measure."
Is an analogous statement true for an arbitrary locally compact topological group $G$ and a Haar measure $v$ on $G$. In other words, does the following hold?
"If $G$ is a locally compact topological group, $v$ a Haar measure on $G$ and $\mu$ is a locally finite measure on $G$ that is left invariant, then there exists a positive constant $a$, such that $\mu=av$."