Autocorrelation for periodic signals
The OP states that for a (deterministic) power signal $x(t)$, the autocorrelation function is defined as
$$R_x(\tau) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt\tag{1}$$ and then wonders whether in the case when $x(t)$ happens to be a periodic signal with period $T_0$, then it is true that
$$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt ??\tag{2}$$
The answer is Yes, for the reasons given below.
Let us consider the integral $\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt$ in $(1)$ when $x(t)$ is periodic with period $T_0$ and $T = nT_0+\alpha$ where $0\leq \alpha &lt; T_0$. Note that the integrand also has period $T_0$.

When $\alpha=0$, the range of integration is over $2n$ periods of the integrand and so
\begin{align}
\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &amp;= \int_{-nT_0}^{nT_0} x(t)x^*(t+\tau) \,\mathrm dt\\
&amp;= 2n\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt.
\end{align}
It follows that
$$\frac{1}{2nT_0}\int_{-nT_0}^{nT_0} x(t)x^*(t+\tau) \,\mathrm dt = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt.$$
The astute reader will have seen that if $\alpha=\frac{T_0}{2}$, the range of integration is over an interval of length $(2n+1)T_0$ (that is, over $2n+1$ periods of the integrand) and so we have
$$\frac{1}{2(n+1)T_0}\int_{-(n+\frac 12)T_0}^{(n+\frac 12)T_0} x(t)x^*(t+\tau) \,\mathrm dt = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt$$
also.
For $\alpha \in \left(0,\frac 12\right)$ (or $\alpha \in \left(\frac 12,1\right)$, the integral is over an integer number of periods (say $N$) of the integrand plus an interval $\beta T_0$ of length less than $\frac 12T_0$ and so
\begin{align}
\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &amp;= N\int_0^{T_0} x(t)x^*(t+\tau)\,\mathrm dt + \Delta
\end{align}
where $\Delta$ denotes whatever the contribution is from the integration of $x(t)x^*(t+\tau)$ over that little interval of length $\beta T_0 &lt; \frac 12 T_0$.
But then we have that
\begin{align}
\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt &amp;= \frac{1}{(N+\beta)T_0}\left[N\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt + \Delta \right]\\
&amp;= \frac{N}{N+\beta}\cdot \left[\frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt\right] + \frac{\Delta}{(N+\beta)T_0}.
\end{align}
But $(1)$ says that we need to take the limit of the above as $T \to \infty$ which we can do by letting $N\to\infty$ and so we see that in all cases,


If $x(t)$ is a deterministic periodic power signal, then its autocorrelation function
$$R_x(\tau) = \lim_{T\to \infty}\frac{1}{2T}\int_{-T}^T x(t)x^*(t+\tau) \,\mathrm dt\tag{1}$$
can be computed via
$$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x^*(t+\tau) \,\mathrm dt, \tag{2}$$ which is what the OP asks for.

For the important special case when $x(t)$ is a real-valued function, Eq. $(2)$ simplifies to
$$R_x(\tau) = \frac{1}{T_0}\int_0^{T_0} x(t)x(t+\tau) \,\mathrm dt, \tag{3}$$
and this is the only case we consider below.

Note that so far, there has been no whiff of Fourier transforms or Fourier series or Dirac deltas or anything to do with the frequency domain, and so let us bring that stuff in.  Note that a real-valued deterministic periodic power signal $x(t)$ can be represented by a Fourier series
$$x(t) = \sum_{n=-\infty}^\infty c_n\exp(j2\pi nf_0t)\tag{4}$$ where $f_0 = T_0^{-1}$ and $c_{-n}=c_n^*$, and does not enjoy a Fourier transform in the classical sense (where Dirac deltas a.k.a. impulses are not allowed). However, $x(t)$ does have a Fourier transform $X(f)$ in the generalized sense where
$$X(f) = \sum_{n=-\infty}^\infty c_n\delta(f-nf_0).\tag{5}$$
In the classical theory, if $X(f)$ is the Fourier transform of $x(t)$, then $|X(f)|^2$ is the Fourier transform of its autocorrelation function $R_x(\tau)$. Does this result hold in the generalized sense too? That is, for a periodic signal $x(t)$ with Fourier series as in $(3)$ and periodic autocorrelation function $R_x(\tau)$ as in $(2)$, what is the Fourier series for $R_x(\tau)$ and the corresponding generalized Fourier transform of $R_x(\tau)$???
We have that $R_x(\tau)$ has Fourier series $\displaystyle \sum_{n=-\infty}^\infty a_n\exp(j2\pi nf_0\tau)$ where
\begin{align}a_n &amp;= \frac{1}{T_0}\int_{0}^{T_0} R_x(\tau)\exp(-j2\pi nf_0\tau)\,\mathrm d\tau\\
&amp;= \frac{1}{T_0}\int_{0}^{T_0} \left[\frac{1}{T_0}\int_{0}^{T_0} x(t)x(t+\tau)\,\mathrm dt\right] \exp(-j2\pi nf_0\tau)\,\mathrm d\tau\\
&amp;= \frac{1}{T_0}\int_{0}^{T_0} x(t)\left[\frac{1}{T_0}\int_{0}^{T_0} x(t+\tau)\exp(-j2\pi nf_0\tau)\,\mathrm d\tau\right] \,\mathrm dt\\
&amp;=\frac{1}{T_0}\int_{0}^{T_0} x(t)\left[\frac{1}{T_0}\int_{0}^{T_0} x(t+\tau)\exp(-j2\pi nf_0(t+\tau))\,\mathrm d\tau\right] \exp(j2\pi nf_0t)\,\mathrm dt\\
&amp;=c_{n}\cdot \frac{1}{T_0}\int_{0}^{T_0} x(t) \exp(j2\pi nf_0t)\,\mathrm dt\\
&amp;= c_n\cdot c_{-n}\\
&amp;= |c_n|^2.
\end{align}
Thus, for a real-valued deterministic periodic power signal $x(t)$,

$$\text{If } x(t) = \sum_{n=-\infty}^\infty c_n\exp(j2\pi nf_0t), ~~\text{then }
R_x(\tau) = \sum_{n=-\infty}^\infty |c_n|^2\exp(j2\pi nf_0\tau). \tag{6}$$

Finally, what about Fourier transforms in the generalized sense?
Well, we have that
\begin{align}
\mathscr F[x(t)] &amp;= \sum_{n=-\infty}^\infty c_n\delta(f-nf_0)\\
\mathscr F[R_x(\tau)] &amp;= \sum_{n=-\infty}^\infty |c_n|^2\delta(f-nf_0)
\end{align}
and so it is not the case that the formula
$$\mathscr F[x(t)] = X(f) \implies \mathscr F[R_x(\tau)] = |X(f)|^2 
\tag{7}$$
of classical Fourier analysis holds unless we are willing to assume that when we multiply out $\left|\sum_{n=-\infty}^\infty c_n\delta(f-nf_0)\right|^2$ into a doubly infinite sum over $m$ and $n$,
$$c_n \delta(f- nf_0)c_m^* \delta^*(f- mf_0) = \begin{cases}
|c_n|^2 \delta(f- nf_0), &amp; \text{if}~m=n,\\
0, &amp;\text{if}~m\neq n,
\end{cases}
\tag{8}$$
holds. In particular, we must treat $|c_n|^2\delta(f-nf_0)\delta^*(f-nf_0)$ as being equal to $|c_n|^2\delta(f-nf_0)$ as stated in my previous answer that the OP has denigrated as invalid and asserted that the claim that, in this context, the square of a Dirac delta should be treated as a Dirac delta, is meaningless because $\delta^2(x)$ is meaningless. The alternative is to insist that $(7)$ holds in classical Fourier analysis only and cannot be applied in the generalized Fourier analysis that allows for Dirac deltas. You pays your money and you takes your choice.....