I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:

First, the commensurator $$ Comm_H(G) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|<\infty\} $$ is a group. We will show:

Lemma: $\varphi\colon Comm_H(G)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.

From the lemma, if we assume there is a $g\in Comm_H(G)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.

Proof of the lemma:

We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

• For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
• $a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

• $a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

• $a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
• $a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
• $b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

• $a a_1'=a_1a_2'$
• $ba_1' = b_2 b_1'$
• $a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get $$ \frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}. $$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$ \varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2). $$

I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:

First, the commensurator $$ Comm_G(H) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|<\infty\} $$ is a group. We will show:

Lemma: $\varphi\colon Comm_G(H)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.

From the lemma, if we assume there is a $g\in Comm_G(H)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.

Proof of the lemma:

We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

• For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
• $a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

• $a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

• $a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
• $a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
• $b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

• $a a_1'=a_1a_2'$
• $ba_1' = b_2 b_1'$
• $a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get $$ \frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}. $$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$ \varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2). $$

I believe (1) and (3) are mutually exclusive. Here is a proof:

First, the commensurator $Comm_H(G)$ is a group. We will show:

Lemma: $\varphi\colon Comm_H(G)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.

From the lemma, if we assume there is a $g\in G$ such that $\varphi(g)=x>1$ (criterion 1), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.

Proof of the lemma:

We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

• For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
• $a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

• $a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

• $a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
• $a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
• $b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

• $a a_1'=a_1a_2'$
• $ba_1' = b_2 b_1'$
• $a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get $$ \frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}. $$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$ \varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2). $$

I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:

First, the commensurator $$ Comm_H(G) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|<\infty\} $$ is a group. We will show:

Lemma: $\varphi\colon Comm_H(G)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.

From the lemma, if we assume there is a $g\in Comm_H(G)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.

Proof of the lemma:

We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:

• For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$
• $a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$

Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:

• $a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$

Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define

• $a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$
• $a_2'=[H\cap g_1Hg_1^{-1}\colon K]$
• $b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$

which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have

• $a a_1'=a_1a_2'$
• $ba_1' = b_2 b_1'$
• $a_2b_1'=b_1a_2'$

Solving for $a$ and $b$, we get $$ \frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}. $$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$ \varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2). $$