The answer is *no*. 
$\mathcal Hom (\mathcal O_X,\omega)\simeq \omega$, but $\omega\otimes \mathcal Hom (\omega, \mathcal O_X)$ is not necessarily reflexive, let alone locally free. However, it is true that 
 if $X$ is $G_1$, that is, Gorenstein in codimension $1$, then
$$
(\omega\otimes \mathcal Hom (\omega, \mathcal O_X))^{**}\simeq \mathcal O_X.
$$

**EDIT:** (due to popular demand, here is a better formed statement for the *CM is not even needed* part of the original)

This is actually true in a little more general setting:
For simplicity assume that $X$ is $G_1$, $S_2$, equidimensional of dimension $d$ and admits a dualizing complex denoted by $\omega_X^\bullet$. (If, say, by *variety* you mean a quasi-projective (reduced) scheme of finite type over a field, then the last assumption is automatic. If in addition you also mean *irreducible*, then so is the equidimensionality. CM obviously implies $S_2$.)

Let
$$\omega_X := h^{-d}(\omega_X^\bullet)$$ 
and 
$$\omega_X^*:=\mathcal Hom_X (\omega_X, \mathcal O_X).$$

Then
$${(\omega_X\otimes \omega_X^*)}^{**}\simeq \mathcal O_X.$$

This follows by the fact that $X$ is $S_2$, both sides are reflexive and they agree in codimension $1$ due to the $G_1$ assumption.

**EDIT2:** (inspired by Karl's answer):
This actually also implies that
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq \mathcal O_X$$
(under the same conditions) since on the open set where $\omega_X$ is a line bundle, 
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq {\omega_X\otimes \omega_X^*}$$
and then since they are both reflexive and $X$ is $S_2$, 
$$\mathcal Hom_X(\omega_X,\omega_X)\simeq ({\omega_X\otimes \omega_X^*})^{**}\simeq \mathcal O_X.$$