When $p$ is an odd prime, is $(p+2)/p$ an outlaw or an index?

Not an answer, just some remarks that are too long to fit in the Comments section.

Notice that, if $I(n)=(p+2)/p$, then $$\frac{n}{D(n)}=\frac{p}{p-2}.$$

Since $p$ is an odd prime, then $\gcd(p,p-2)=1$. Thus, $D(n) \nmid n$, unless $p=3$.

This implies that if $I(n)=(p+2)/p=5/3$ (when $p=3$), then $n$ is deficient-perfect.

Otherwise, if $p>3$, then since $p$ is an odd prime, $p \geq 5$, so that $$\frac{n}{D(n)}=\frac{p}{p-2}=\frac{1}{1-\frac{2}{p}} \leq \frac{1}{1-\frac{2}{5}}=\frac{1}{\frac{3}{5}}=\frac{5}{3}$$ and $$\frac{n}{D(n)}=\frac{(p-2)+2}{p-2}=1+\frac{2}{p-2}>1,$$ from which we obtain $$1 < \frac{n}{D(n)} \leq \frac{5}{3},$$ which implies that $D(n) \nmid n$ for $p>3$.

Since $$1 < \frac{n}{D(n)} \leq \frac{5}{3}$$ implies that $$1 < I(n) \leq \frac{7}{5},$$ and since we have $n$ is a square if $I(n)=(p+2)/p$ and $p$ is an odd prime, and because $$\frac{8}{5} < I(m^2) < 2$$ if $q^k m^2$ is an odd perfect number with Euler prime $q$, then we have that $$I(m^2)=\frac{p+2}{p} \iff p=3.$$

Too long to comment.

One can prove that

$$\frac{2n}{n+D(n)}<\frac{(2a+2)n−aD(n)}{(a+1)n+D(n)}<I(n)\tag1$$

$$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}\tag2$$

hold for any $a>0,b\gt -1$. Note here that $a,b$ are not necessarily integers.

However, it seems that we cannot prove the conjecture using $(1)(2)$.

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About $(1)$ :

Let $x:=\frac{\sigma(n)}{n}$. Then, we get $$1\lt x\lt 2\tag3$$

Trying to find $a,b,c$ such that $-x+c\gt 0$ and $$\frac{2}{3-x}\lt\frac{ax+b}{-x+c}\lt x$$ which is equivalent to $$ax^2+(b-3a-2)x+2c-3b\lt 0\quad\text{and}\quad x^2+(a-c)x+b\lt 0\tag4$$

every $x$ such that $(4)$ has to satisfy $(3)$.

So, trying to find $a,b,c$ such that

$$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(a-c)^2-4b}\le \min(4+a-c,-a+c-2)\\\sqrt{(3a+2-b)^2-4a(2c-3b)}\le \min(a+2-b,a-2+b)\end{cases}$$ and choosing $b=2$ give $$\begin{cases}a\gt 0\\c\ge 2\\\sqrt{(c-a)^2-8}\le \min(4-(c-a),(c-a)-2)\\\sqrt{9a^2-8a(c-3)}\le a\end{cases}$$

So, we see that choosing $c=a+3$ works, and that $$\frac{2}{3-x}\lt\frac{2+ax}{a+3-x}\lt x,$$ i.e. $$\frac{2n}{n+D}\lt\frac{(2a+2)n-aD}{(a+1)n+D}\lt \frac{\sigma(n)}{n}$$ holds for any $a\gt 0$.

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About $(2)$ :

Trying to find $a,b,c$ such that $-x+c\gt 0$ and $$x\lt\frac{a+bx}{-x+c}\lt\frac{4-x}{3-x}$$ which is equivalent to $$x^2+(b-c)x+a\gt 0\quad\text{and}\quad (b+1)x^2+(a-3b-c-4)x+4c-3a\gt 0\tag6$$ every $x$ such that $(6)$ has to satisfy $x\not=2$.

So, trying to find $a,b,c$ such that

$$\begin{cases}b+1\gt 0\\ (b-c)^2-4a\le 0\\ (a-3b-c-4)^2-4(b+1)(4c-3a)\le 0\\ 2^2+(b-c)\times 2+a=0\\ (b+1)\times 2^2+(a-3b-c-4)\times 2+4c-3a=0\end{cases}$$

and choosing $a=4$ give $$\begin{cases}b\gt -1\\ c=b+4\end{cases}$$

So, we see that $$x\lt\frac{4+bx}{-x+b+4}\lt\frac{4-x}{3-x},$$ i.e. $$I(n)\lt\frac{(2b+4)n-bD(n)}{(b+2)n+D(n)}\lt \frac{2n+D(n)}{n+D(n)}$$ holds for any $b\gt -1$.