Let $T : X \to X$ be a measurable map and let $P =\{a_1, \cdots ,a_n\}$ be a periodic orbit with minimal period $n$. That is, $T(a_i) = a_{i+1}$ for $i = 1, \cdots ,n$ and $T(a_n) = a_1$. Let $p_1 , \cdots , p_n$ be constants such that $p_i \in (0,1)$ and $\Sigma_{i=1}^n p_i=1$. Consider the measure $\delta_p(A):=\Sigma_{\{i;p_i \in A\}}p_i$
Now I want to prove that $\delta_p$ is invariant iff $\forall i \quad p_i=\frac{1}{n}$
If $\delta_p$ is invariant under $T$ then $\delta_p(A)=\delta_p(T^{-1}(A))$ so $\Sigma_{\{i;p_i \in A\}}p_i =\Sigma_{\{i;p_i \in T^{-1}A\}}p_i $.
I don't know how to reach the fact that $p_i$ must be $\frac{1}{n}$ for all $i=1 , \cdots, n$
Hint: We can write $\delta_{p} = \sum_{i = 1}^{n} p_{i} \delta_{a_{i}}$, where $\delta_{a_{i}}$ is the measure such that $\delta_{a_{i}}(A) = 1$ if $a_{i} \in A$ and $\delta_{a_{i}}(A) = 0$, otherwise. Here we interpret the sum of measures pointwise. That means \begin{equation*} \delta_{p}(A) = \sum_{i=1}^{n} p_{i} \delta_{a_{i}}(A). \end{equation*}
Hint: $\delta_{p}(T^{-1}(A)) = \sum_{i = 1}^{n} p_{i} \delta_{a_{i}}(T^{-1}(A))$.
Hint: In general, if $x \in X$, then $\delta_{x}(T^{-1}(A)) = \delta_{T(x)}(A)$.
The previous two hints are related to fundamental properties of the so-called pushforward of a measure: it is linear and it has certain covariance/contravariance properties. Pushforward is a super useful operation; it's worth doing a little digging in textbooks to learn the basic properties/results (e.g. change of measure).