so let's say the length of your FFT is $N$. let's say that you fill half of the buffer with your signal and fill the other half with zeros.
$$ x[n] = 0 \quad \text{ for } \tfrac{N}2 \le n < N $$
the DFT of $x[n]$ is $X[k]$. then magnitude-square and inverse DFT:
$$ R_x[n] = \mathcal{iDFT}\bigg\{ \bigg|X[k]\bigg|^2 \bigg\} $$
in the time domain, this comes out as
$$ R_x[\ell] = \sum\limits_{n=0}^{\tfrac{N}2 - 1 - \ell} x[n] x[n+\ell] $$
we call $\ell$ the "lag" of the autocorrelation.
now suppose $x[n]$ is periodic with some period $P$ which is much smaller than $\frac{N}2$.
$$ x[n+P] = x[n] \quad \text{for } 0 \le n < n+P < \tfrac{N}2 $$
with a lag of zero
$$ R_x[0] = \sum\limits_{n=0}^{\tfrac{N}2 - 1} (x[n])^2 $$
which, if we apply a little bit of statistical sleight-of-hand, we might expect is $\tfrac{N}2$ times the expected power $\overline{x^2}$. in the limit with $P \ll \tfrac{N}2 $we can sorta do this with periodic $x[n]$.
$$\begin{align}
R_x[0] &= \sum\limits_{n=0}^{\tfrac{N}2 - 1} (x[n])^2 \\
&\approx \sum\limits_{n=0}^{\tfrac{N}2 - 1} \overline{x^2} \\
&= \frac{N}2 \overline{x^2}
\end{align} $$
now with a lag of $mP$ where $m>0$ is an integer:
$$\begin{align}
R_x[mP] &= \sum\limits_{n=0}^{\tfrac{N}2 - 1 - mP} x[n] x[n+mP] \\
&= \sum\limits_{n=0}^{\tfrac{N}2 - 1 - mP} x[n] x[n] \\
&= \sum\limits_{n=0}^{\tfrac{N}2 - 1 - mP} (x[n])^2 \\
&\approx \sum\limits_{n=0}^{\tfrac{N}2-1 - mP} \overline{x^2} \\
&= \left( \frac{N}2 - mP \right) \overline{x^2} \\
\end{align} $$
eliminating $\overline{x^2}$ we have
$$ R_x[mP] \approx \left( 1 - \frac{2mP}N \right) R_x[0] $$
for $P \ll \tfrac{N}2$.
this suggests an envelope on the autocorrelation of
$$ \big| R_x[\ell] \big| \le \left( 1 - \frac{2 |\ell|}N \right) R_x[0] $$
this is for the rectangular window. it can be generalized for other windows. the envelope results from the two windows having less area in common as the lag $\ell$ increases.