If $X$ has no finite dimensional invariant subspaces (other than $\{0\}$) then the only invariant probability measure is the Dirac measure at $0$.
More generally, let $X_0$ be the closure in $X$ of the vector space generated by all invariant finite dimensional subspaces. Then every invariant probability measure $\mu$ on $X$ is supported on $X_0$ and it is given as follows:

Fix a probability measure $\nu$ on the orbit space $X_0/G$, where $G$ is the Bohr compactification of $\Gamma$ (recall that the $\Gamma$ action on $X_0$ extends to $G$) and set
$$\mu=\int_{X_0/G} (\text{The normalized Haar measure on the orbit})~ d\nu(\text{orbit}).$$
 
In particular, every ergodic $\Gamma$-measure is supported (on $X_0$ and) on a unique $G$-orbit, and it is the Haar measure on it.
Note that by ergodic decomposition it is enough to prove the last statement.

Claim: let $\Gamma$ be a group acting isometrically on a metric space $Y$ preserving an ergodic probability measure $\mu$. Assume $Y$ is complete and $\mu$ is fully supported. Then there is a $\Gamma$-equivariant homeomorphism $Y\simeq G/L$, where $G$ is the Bohr compactification of $\Gamma$ and $L<G$ a closed subgroup, and under this identification $\mu$ correspond to the Haar measure.

The essential part of the claim is the compactness of $Y$, which follows by completeness from total-boundedness, which is easy (generalize the standard claim that $Y$ is bounded, which is proven by taking a compact set of more than half the weight). Then the isometry group of $Y$ is compact by Arzela-Ascoli and the $\Gamma$ action extends to $G$. By ergodicity (and the fact that $Y/G$ is nice) we get that $\mu$ is supported on a unique orbit and everything else follows easily.

Now, for the statement we had regarding ergodic measures, take $Y$ to be the support of an ergodic measure on $X$, which I assume to be total. Observe that the action of $G$ on $Y$ extends to a linear isometric action on $X$. Call Peter-Weyl to see that $X=X_0$.