To answer question 1, I think there will always be a measure for which $T$ is ergodic.

Rotation of the unit circle $T(z) = az$ is measure preserving, (with Haar measure) and ergodic only when $a$ is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8).

However, we can still define a measure for which $T$ is ergodic. Say $\mu(\{a^k\}) = 1/n$ where $a$ an $n$-th root of unity, $k=0,\ldots, n-1$. Then the $T$-invariant sets either have measure 0 or 1.

When there are periodic points we can use the above argument to create an ergodic measure.

The argument is not very different in the absence of periodic points. For any measure preserving system $(X,\mathcal B, \nu)$ take any element $a \in X$ and define $orb_T(a) = \{T^na: n \in \mathbb Z\}$. Let $\mu$ be a probability measure on $\mathcal B / orb_T(a)$ with $\mu(0 + orb_T(a)) = 1$ and zero otherwise. Then $T$ is ergodic on $(X, \mathcal B/ orb_T(a), \mu)$.

I'd like to improve upon my response. I read it again after 4 years and I did not recognise myself.

First, I'd like to point out that asymptotic density is an ergodic and $T$-invariant probability measure on the set of integers $\mathbb Z$ with $T(x) = x+1$.

I'll withdraw the rest of my comment.

To answer question 1, I think there will always be a measure for which $T$ is ergodic.

Rotation of the unit circle $T(z) = az$ is measure preserving, (with Haar measure) and ergodic only when $a$ is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8).

However, we can still define a measure for which $T$ is ergodic. Say $\mu(\{a^k\}) = 1/n$ where $a$ an $n$-th root of unity, $k=0,\ldots, n-1$. Then the $T$-invariant sets either have measure 0 or 1.

When there are periodic points we can use the above argument to create an ergodic measure.

The argument is not very different in the absence of periodic points. For any measure preserving system take any element $a \in X$ and define a new measure $\mu(T^ka) = 3.2^{-|k|-1}$. The $T$-invariant sets either contain the orbit of $a$ and have measure 1; or are disjoint from the orbit of $a$ and have measure 0.

To answer question 1, I think there will always be a measure for which $T$ is ergodic.

Rotation of the unit circle $T(z) = az$ is measure preserving, (with Haar measure) and ergodic only when $a$ is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8).

However, we can still define a measure for which $T$ is ergodic. Say $\mu(\{a^k\}) = 1/n$ where $a$ an $n$-th root of unity, $k=0,\ldots, n-1$. Then the $T$-invariant sets either have measure 0 or 1.

When there are periodic points we can use the above argument to create an ergodic measure.

The argument is not very different in the absence of periodic points. For any measure preserving system $(X,\mathcal B, \nu)$ take any element $a \in X$ and define $orb_T(a) = \{T^na: n \in \mathbb Z\}$. Let $\mu$ be a probability measure on $\mathcal B / orb_T(a)$ with $\mu(0 + orb_T(a)) = 1$ and zero otherwise. Then $T$ is ergodic on $(X, \mathcal B/ orb_T(a), \mu)$.