Is Haar measure reflection invariant?

Let $$G$$ be a locally compact group and $$\mu$$ be a Haar measure on $$G.$$ Is $$\mu$$ necessarily reflection invariant i.e. can we always say that $$\mu (E) = \mu (E^{-1})\$$?

where $$E \subseteq G$$ is a measurable set and $$E^{-1} = \left \{x \in G\ |\ x^{-1} \in E \right \}.$$

For Lebesgue measure I know that the result holds. But for Haar measure defined on an arbitrary locally compact group I don't know the result.

Your assumption means that a left Haar measure is also a right Haar measure, so your assumption happens if and only if the locally compact group $$G$$ is unimodular.
So, for a concrete counterexample, consider any non-unimodular group such as the $$ax+b$$-group.
Note that, if $$\chi_E$$ is the characteristic function of $$E$$, then$$\mu(E)=\int_G\chi_E\,\mathrm d\mu\quad\text{and}\quad\mu\left(E^{-1}\right)=\int_G\chi_E\circ\iota\,\mathrm d\mu,$$with $$\iota(x)=x^{-1}$$. But, in general, we do not have$$\int_Gf(x)\,\mathrm d\mu(x)=\int_Gf(x^{-1})\,\mathrm d\mu(x);$$this happens only for the unimodular groups.
If you want a concrete example, take$$G=\left\{\begin{bmatrix}a&b\\0&1\end{bmatrix}\,\middle|\,a>0\wedge b\in\Bbb R\right\}.$$A left Haar measure for this group is$$\mu(E)=\iint_E\frac1{x^2}\chi_E\,\mathrm dx\,\mathrm dy.$$Now, take $$E=[1,2]\times[0,1].$$ Then $$\mu(E)=\frac12$$, but $$\mu\left(E^{-1}\right)=\log(2)$$.