Let $X= (X_t)_{t \in \mathbb Z}$ be a strictly stationary process with $E[X_t]=0$ for all $t$. We know that the autocorrelation functions is given by
$$\gamma_X(h)= E[X_0 X_h ],\quad h > 0$$
Now consider $X_1= (X_{t;1})_{t \in \mathbb Z}$, $X_2= (X_{t;2})_{t \in \mathbb Z}$, $X_3= (X_{t;3})_{t \in \mathbb Z}$,… infinite copies of $X= (X_t)_{t \in \mathbb Z}$. Copies of $X= (X_t)_{t \in \mathbb Z}$ means that $\{X_{j}\}_{j=1}^{\infty}$, where $X_j = (X_{t;j})_{t \in \mathbb Z}$ , is i.i.d. acording to $X=(X_t)_{t \in \mathbb Z}$. So, consider de following random sum:
$$Y_t = \sum_{j=1}^N X_{t;j}, \quad N \sim \hbox{Poisson}(\lambda)$$
Note that this is not a Compound Poisson random process, because $N$ is fixed and does not depend on $t$. Besides this, $\{X_{j}\}_{j=1}^{\infty}$ are independent of $N$.
Note that $(Y_t)_{t \in \mathbb Z}$ is stationary and
$$E[Y_t]= \lambda E[X_t]=0, \quad \forall t \in \mathbb Z$$
So I want to find the Autocovariance Function of $Y=(Y_t)_{t \in \mathbb Z}$:
$$\gamma_Y(h)= E[Y_0 Y_h ]$$
in terms of $\gamma_X(h)$, the autocorrelation function of $X= (X_t)_{t \in \mathbb Z}$.
Best Answer
\begin{align} E\big(Y_0Y_h\,\big|\,N\big)&=\sum_{j=1}^N\sum_{k=1}^NE\big(X_{0;j}X_{h;k}\big)\\ &=\sum_{j=1}^NE\big(X_{0;j}X_{h;j}\big)\\ &=N\gamma_X(h)\ . \end{align} Therefore \begin{align} E\big(Y_0Y_h)&=E\big(E\big(Y_0Y_h\,\big|\,N\big)\big)\\ &=E\big(N\gamma_X(h)\big)\\ &=\lambda\gamma_X(h) \end{align}