One has $\max\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments.

We also have $\max\big\{\mu\big(\text{shr}(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$. To see why, define a function $\varphi:\mathbb{N}\to\mathbb{N}$ in the following way. Consider a sequence of intervals $I_n=\{a_n,\dots,a_{n}+n-1\}$ in $\mathbb{Z}$, where $a_1=1$ and $a_{n+1}=a_n+n+2$, so that there are gaps $2$ points between consecutive intervals.

Also let $J_n=\{b_n,\dots,b_{n}+n-1\}$, where $b_1=1$ and $b_{n+1}=b_n+n+1$.

Note that the sets $\cup_nI_n$ and $\cup_nJ_n$ have density $1$ in $\mathbb{N}$.

Partially define a map $\varphi:\mathbb{N}\to\mathbb{N}$ by $\varphi(a_n+k)=b_n+k$ for all $k=1,\dots,n-1$. So $\varphi$ maps $\cup_nI_n$ bijectively to $\cup_nJ_n$. Extend $\varphi$ to a bijection in $\mathbb{N}$, it doesn't matter how.

Note that for any $a\in I_n$, $b\in I_m$ with $n\neq m$, we have $|\varphi(a)-\varphi(b)|<|a-b|$. This implies that $\mu_{[\mathbb{N}]^2}(\text{shr}(\varphi))=1$, as we wanted.

One has $\max\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments.

We can also create a function $\varphi$ with $\mu\big(\text{shr}(\varphi)\big)=1$ in the following way. Consider a sequence of intervals $I_n=\{a_n,\dots,a_{n}+n-1\}$ in $\mathbb{Z}$, where $a_1=1$ and $a_{n+1}=a_n+n+2$, so that there are gaps $2$ points between consecutive intervals.

Also let $J_n=\{b_n,\dots,b_{n}+n-1\}$, where $b_1=1$ and $b_{n+1}=b_n+n+1$.

Note that the sets $\cup_nI_n$ and $\cup_nJ_n$ have density $1$ in $\mathbb{N}$.

Partially define a map $\varphi:\mathbb{N}\to\mathbb{N}$ by $\varphi(a_n+k)=b_n+k$ for all $k=1,\dots,n-1$. So $\varphi$ maps $\cup_nI_n$ bijectively to $\cup_nJ_n$. Extend $\varphi$ to a bijection in $\mathbb{N}$, it doesn't matter how.

Note that for any $a\in I_n$, $b\in I_m$ with $n\neq m$, we have $|\varphi(a)-\varphi(b)|<|a-b|$. This implies that $\mu_{[\mathbb{N}]^2}(\text{shr}(\varphi))=1$, as we wanted.

One has $\sup\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments.

We also have $\sup\big\{\mu\big(\text{shr}(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$. To see why, define a function $\varphi:\mathbb{N}\to\mathbb{N}$ in the following way. Consider a sequence of intervals $I_n=\{a_n,\dots,a_{n}+n-1\}$ in $\mathbb{Z}$, where $a_1=1$ and $a_{n+1}=a_n+n+2$, so that there are gaps $2$ points between consecutive intervals.

Also let $J_n=\{b_n,\dots,b_{n}+n-1\}$, where $b_1=1$ and $b_{n+1}=b_n+n+1$.

Note that the sets $\cup_nI_n$ and $\cup_nJ_n$ have density $1$ in $\mathbb{N}$.

Partially define a map $\varphi:\mathbb{N}\to\mathbb{N}$ by $\varphi(a_n+k)=b_n+k$ for all $k=1,\dots,n-1$. So $\varphi$ maps $\cup_nI_n$ bijectively to $\cup_nJ_n$. Extend $\varphi$ to a bijection in $\mathbb{N}$, it doesn't matter how.

Note that for any $a\in I_n$, $b\in I_m$ with $n\neq m$, we have $|\varphi(a)-\varphi(b)|<|a-b|$. This implies that $\mu_{[\mathbb{N}]^2}(\text{shr}(\varphi))=1$, as we wanted.

One has $\max\big\{\mu\big(\exp(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$, as mentioned by Emil Jeřábek in the comments.

We also have $\max\big\{\mu\big(\text{shr}(\varphi)\big) :\varphi \in S_{\mathbb{N}}\big\}=1$. To see why, define a function $\varphi:\mathbb{N}\to\mathbb{N}$ in the following way. Consider a sequence of intervals $I_n=\{a_n,\dots,a_{n}+n-1\}$ in $\mathbb{Z}$, where $a_1=1$ and $a_{n+1}=a_n+n+2$, so that there are gaps $2$ points between consecutive intervals.

Also let $J_n=\{b_n,\dots,b_{n}+n-1\}$, where $b_1=1$ and $b_{n+1}=b_n+n+1$.

Note that the sets $\cup_nI_n$ and $\cup_nJ_n$ have density $1$ in $\mathbb{N}$.

Partially define a map $\varphi:\mathbb{N}\to\mathbb{N}$ by $\varphi(a_n+k)=b_n+k$ for all $k=1,\dots,n-1$. So $\varphi$ maps $\cup_nI_n$ bijectively to $\cup_nJ_n$. Extend $\varphi$ to a bijection in $\mathbb{N}$, it doesn't matter how.

Note that for any $a\in I_n$, $b\in I_m$ with $n\neq m$, we have $|\varphi(a)-\varphi(b)|<|a-b|$. This implies that $\mu_{[\mathbb{N}]^2}(\text{shr}(\varphi))=1$, as we wanted.