Yes. This follows from the fact that $G_2$ acts on the octonions via automorphisms.
Let $e_i$, $i=1,\dots,7$ be a choice of 7 imaginary octonion units and let $1$ denote the identity. Then by definition of the $O_i$, we have that
$$
e_i e_j = \sum_k e_k (O_i)_{kj} - \delta_{ij} 1
$$
where the left-hand side is octonion multiplication.
Let $g \in G_2$. Since $G_2$ acts on the octonions via automorphisms,
$$
g(e_i) = \sum_j e_j M_{ji}~.
$$
The $M_{ji}$ are the entries of the matrix $M_\gamma$ in the question.
Now apply $g \in G_2$ on the first displayed equation. On the LHS we have
$$
g(e_i e_j) = g(e_i) g(e_j) = \sum_{k,l} \left(\sum_m e_m (O_k)_{ml} - \delta_{kl} 1\right) M_{ki} M_{lj} = \sum_k (O_k M)_{mj} M_{ki} - \delta_{ij} 1~,
$$
where we have used that $M$ is orthogonal, so that $\sum_{k,l} \delta_{kl} M_{ki} M_{lj} = \delta_{ij}$. Similarly, on the RHS we find
$$
\sum_{m,k} e_m M_{mk} (O_i)_{kj} - \delta_{ij} 1
$$
using that $g(1) = 1$.
Comparing the two, we find
$$
\sum_{k,l,m} e_m (O_k)_{ml} M_{ki} M_{lj} = \sum_{m,k} e_m M_{mk} (O_i)_{kj}
$$
which is equivalent to
$$
\sum_k (O_k M)_{mj} M_{ki} = (M O_i)_{mj}
$$
or, finally,
$$
M O_i M^{-1} = \sum_k O_k M_{ki}~.
$$
Edit To answer the comment, there is a coordinate-free proof of this fact, and perhaps I should have written that first. What is $M O_i M^{-1}$? It is the matrix of left multiplication by the imaginary octonions (projected to the imaginary octonions) under a $G_2$ change of basis. But $G_2$ is an automorphism of the octonions, hence it preserves the subspace of imaginary octonions. Hence $M O_i M^{-1}$ is the matrix of left multiplication by some imaginary octonion unit, whence some linear combination of the $O_i$, since these form a basis.
