We have the following:
$m=8=4\cdot 2$ :
$1\cdot 3\cdot 5\cdot 7 \equiv 1\cdot 3\cdot (-3)\cdot (-1) \equiv (-1^2)\cdot (-3^2)\equiv (-1) \cdot (-1) \equiv 1$
$m=10=2\cdot 5$ :
$1\cdot 3\cdot 7\cdot 9 \equiv 1\cdot 3\cdot 3^{-1}\cdot(-1) \equiv -1^2\cdot (3\cdot 3^{-1}) \equiv (-1) \cdot 1 \equiv -1$
$m=12=4\cdot 3$ :
$1\cdot 5\cdot 7\cdot 11 \equiv 1\cdot 5\cdot (-5)\cdot (-1) \equiv (-1) \cdot (-1) \equiv 1$
$m=14=2\cdot 7$ :
$1\cdot 3\cdot 5\cdot 9\cdot 11\cdot 13 \equiv 1\cdot 3\cdot 3^{-1}\cdot 9\cdot 9^{-1}\cdot (-1) \equiv (-1^2)\cdot (3\cdot 3^{-1})\cdot (9\cdot 9^{-1})\equiv (-1)\cdot 1\cdot 1\equiv -1 $
$m=16=4\cdot 4$ :
$1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11\cdot 13\cdot 15 \equiv 1\cdot 3\cdot 5\cdot 7\cdot (-7)\cdot 3^{-1}\cdot 5^{-1}\cdot (-1) \equiv (-1^2) \cdot (3\cdot 3^{-1})\cdot (5\cdot 5^{-1})\cdot (7\cdot (-7))\equiv (-1)\cdot 1\cdot 1\cdot (-1)\equiv 1$
Does it hold in the general case that when $m$ is a multiple of $4$ the result is $1$ and when $m$ is a multiple of $2$ but not of $4$ the result is $-1$?
Let's check.
For $m=4$ we have $1\cdot 3\equiv 1 \cdot -1 \equiv -1$.
No, I don't think it holds.
Still, isn't there a pattern that we can either pair a number with its multiplicative inverse, or we can pair it with its additive inverse?
Is that always possible?
And can we always tell what the product will be of such a pair?
I have thought the following:
We want to examine the product $B=b_1 b_2 \cdots b_{\phi(m)}$.
If $\exists j \in \{2, \ldots , \phi(m)-1\}$ such that $b_j^{-1}\not\equiv b_j\mod m$ then at the product $b_2\cdot \ldots \cdot b_{\phi(m)-1}$ there is a pair such that their product is equal to $1$.
Let $b_k$ be the modular inverse of $b_j$, i.e., $b_j^{-1}\equiv b_k\mod m$.
We get that \begin{align*}b_2\cdot b_3 \cdot \ldots \cdot b_j\cdot \ldots b_k\cdot \ldots \cdot b_{\phi(m)-1}&\equiv b_2\cdot b_3 \cdot \ldots \cdot b_j\cdot \ldots b_j^{-1}\cdot \ldots \cdot b_{\phi(m)-1} \\ & \equiv b_2\cdot b_3 \cdot \ldots \cdot (b_j\cdot b_j^{-1})\cdot \ldots \cdot b_{\phi(m)-1} \\ & \equiv b_2\cdot b_3 \cdot \ldots \cdot 1\cdot \ldots \cdot b_{\phi(m)-1}\end{align*}
If $\exists r\in \{2, \ldots , \phi(m)-1\}$ such that $b_r^{-1}\equiv b_r\mod m$ then we have that $b_r^2\equiv 1\mod m$.
In this case we cannot find two different factors of the product, $b_i$ and $b_r$ with $i\neq r$, such that their product is equal to $1$.
We define $n:=m-b_r$ and we get $n\equiv -b_r\mod m$.
It holds that $(n,m)=(-b_r,m)=(b_r,m)=1$, i.e., $n$ and $m$ are comprime, so $n$ is one of the $b_i$'s with $i\neq r$, i.e., $n:=b_i$.
So, we get that
\begin{align*}b_2\cdot b_3 \cdot \ldots \cdot b_r\cdot \ldots b_i\cdot \ldots \cdot b_{\phi(m)-1}&\equiv b_2\cdot b_3 \cdot \ldots \cdot b_r\cdot \ldots (-b_r)\cdot \ldots \cdot b_{\phi(m)-1} \\ & \equiv b_2\cdot b_3 \cdot \ldots \cdot (-b_r^2)\cdot \ldots \cdot b_{\phi(m)-1} \\ & \equiv b_2\cdot b_3 \cdot \ldots \cdot (-1)\cdot \ldots \cdot b_{\phi(m)-1}\end{align*}
Continuing in the same way , the product $b_2 b_3 \cdots b_{\phi(m)-1}$ will be equal either to $1$ or to $-1$.
So $B=1 \cdot b_2 \cdots b_{\phi(m)-1} (m-1) \equiv -1 \cdot b_2 \cdots b_{\phi(m)-1} \equiv \pm 1 \pmod{m}$.
Am I right?
Erm... this is bit much...
Starting somewhere, what happened to $b_1$ and $b_{\phi(m)}$?