Autocorrelation Jitter
The (periodic) autocorrelation function of a sinusoid $A\cos(2\pi f_0 t + \theta)$
is 

$$\begin{align}
R(\tau) &amp;= \int_0^{f_0^{-1}} A\cos(2\pi f_0 t + \theta)
A\cos(2\pi f_0 (t +\tau)+ \theta)\,\mathrm dt\\
&amp;= \frac{A^2}{2}\int_0^{f_0^{-1}} \cos(2\pi f_0\tau)+\cos(2\pi f_0 (2t +\tau)+ \theta)\,\mathrm dt\\
&amp;= \frac{A^2}{2f_0}\cos(2\pi f_0\tau)
\end{align}$$
which has a peak at $\tau = 0$. Notice in
particular that there is no dependence on $\theta$.   More generally,
if the integration is carried out over an interval $[0,T]$
of arbitrary length, we get
$$\begin{align}
\hat{R}(\tau) &amp;= \int_0^{T} A\cos(2\pi f_0 t + \theta)
A\cos(2\pi f_0 (t +\tau)+ \theta)\,\mathrm dt\\
&amp;= \frac{A^2}{2}\int_0^{T} \cos(2\pi f_0\tau)+\cos(2\pi f_0 (2t +\tau)+ \theta)\,\mathrm dt\\
&amp;= \frac{A^2T}{2}\cos(2\pi f_0\tau) + \left.\frac{A^2}{2}\frac{\sin(2\pi f_0 (2t +\tau)+ \theta)}{4\pi f_0}\right|_0^T\\
&amp;= \frac{A^2T}{2}\cos(2\pi f_0\tau) + \frac{A^2}{2}\frac{\sin(2\pi f_0 (2T +\tau)+ \theta)-\sin(2\pi f_0\tau + \theta)}{4\pi f_0}\\
&amp;= \frac{A^2T}{2} \left[\cos(2\pi f_0\tau) + \frac{\sin(2\pi f_0 (2T +\tau)+ \theta)-\sin(2\pi f_0\tau + \theta)}{4\pi f_0T}\right]
\end{align}$$
which does depend on $\theta$ as well as $T$. Notice that 
if $T$ is an integer multiple $kf_0^{-1}$ of the fundamental
period $f_0^{-1}$, then the second term
in brackets disappears entirely and we get
$$\hat{R}(\tau) = kR(\tau).$$
On the other hand, since the numerator in that second term cannot
have value more than $2$, we get that if $T \gg f_0^{-1}$ but
is not an integer multiple of $f_0^{-1}$, the
second term is quite small in magnitude and so the autocorrelation
function that one deduces from $\hat{R}(\tau)$ depends only
very slightly on $\theta$.

With that as prologue, I point out to you that the FFT method of 
computing the autocorrelation function of a signal gives the
periodic autocorrelation function of a segment of the
signal, that is, what one gets is effectively an approximation to 
$\hat{R}_x(\tau)$ for a periodic signal $x(t)$ that is approximately 
$2.5$ periods of a sinusoid repeated over and over again.  In particular,
this periodic repetition is not the same as the sinusoid continuing
smoothly and without break as one would want it to be for the above
results to apply.  Thus, it is not at all surprising that the
results from the 
FFT method of computing the autocorrelation function depend more
heavily on the phase of the sinusoid. In approximately
$2.5$ periods, the $sin(\cdot)$ goes from $0$ at $t=0$ to nearly
$0$ again at the end of nearly $2.5$ periods. Thus, the periodic
extension is nearly continuous (though there is an abrupt
change of phase at the transition). On the other hand, with
a $90$-degree phase shift, $\cos(\cdot)$ goes from $1$ at 
$t = 0$ to nearly $-1$ at the end of nearly $2.5$ periods,
and so the periodic extension is very discontinuous. This
shows up in the FFT results which don't get a good estimate of
the frequency in the latter case.

If you take the frequency of the sinusoid to be be such
that an integer number of cycles fit into the $N$ that
you are using for your FFT, these artifacts will disappear.