Isomorphism between $L$-structures I have a question in particular on b), since this should simply be wrong, and only true for $n=m$. Also I have a concern on d).
Please have a look. I know it is a lot of text... Thank you!

Let $L$ be the language $\{c\}$ where $c$ is constant symbol. Let $L^<=\{c,<\}$, $L^+=\{c,+\}$, $L^{<,+}=\{c,<,+\}$, where $+$ is a 2-digit function symbol and $<$ is a 2-digit relation symbol.
We observe the structure $S_n$ with $\mathbb{Z}$ as universe (basis set) and $c$ gets interpreted as $n$.
A $S_n$ structure can be expanded to a $L^{+,<}$-structure $S_n^{<,+}$, where $+,<$ have the usual interpretation. Similar $S_n^<$ and $S_n^+$.
Show:
a) $S_n\cong S_m$ for every $n,m\in\mathbb{Z}$
b) For every $n,m\in\mathbb{Z}$ exists exactly one isomorphism between $S_n^<$ and $S_m^<$.
c) $S_n^{<,+}\cong S_m^{<,+}\iff n=m$
d) When does $S_n^+\cong S_m^+$ hold?

a) is simple. We define $f:\mathbb{Z}\to\mathbb{Z}, f(x)=\begin{cases} m,\text{if x=n}\\ n,\text{if x=m}\\ x,\text{else}\end{cases}$
Obviously $f$ is a bijection and it holds $f(c^{S_n})=f(n)=m=c^{S_m}$. So $S_n\cong S_m$.
But b) should only hold for $n=m$, because we have to define the isomorphism as in a). Else we do not map constant symbols onto constant symbols.
And then $n<m\Leftrightarrow f(n)<f(m)\Leftrightarrow m<n$ which is a contradiction.
You can only make sense out of this when $n=m$, or am I missing something?
For c): $\Leftarrow$ is clear.
$\Rightarrow$. Let $S_n^{<,+}\cong S_m^{<,+}$ suppose $n\neq m$. Without loss of generality $n<m$. Then as in b) $n<m\Leftrightarrow m<n$ contradiction.
d):
This can not hold if $n\neq m$ and $n=0$ or $m=0$. Let $n=0$ (so $m\neq 0$).
Then $f(0)=f(0+0)\Leftrightarrow m=2m\Leftrightarrow m=0$ contradiction.
In particular this can not hold if $n\neq m$.
Let $n\cdot m:=\underbrace{m+\dotso +m}_{\text{n-times}}$ if $n>0$ and
$n\cdot m:=\underbrace{(-m)+\dotso (-m)}_{\text{n-times}}$ if $n<0$.
We have $f(n\cdot m)=f(m\cdot n)\Leftrightarrow n\cdot f(m)=m\cdot f(n)\Leftrightarrow n^2=m^2\Leftrightarrow n=m \vee n=-m$.
But $n\neq m$. So $n=-m$.
Then $f(n)=m=-n$. Choose $k\in\mathbb{Z}$ such that $k\neq m,n$ and $n-k\neq m$. Then $f(n)=f((n+(-k))+k)=(n+(-k))+k=n$. So $n=-n$ but $n\neq 0$.
Is this correct? This seems overly 'complicated'.
Thanks in advance.
 A: b) You are missing something. Instead of just swapping $n$ and $m,$ define the isomorphism as $x\mapsto x +(m-n).$ Will leave it to you to show this is order-preserving and the only possibility.
c) Similar error to (b). We know what the order isomorphism has to be from the uniqueness proved in (b) and it's pretty easy to see this doesn't preserve addition unless $m=n.$
d) Any addition-preserving map $f:\mathbb Z\to \mathbb Z$ is determined entirely by $f(1).$ What can $f(1)$ be in order that the map be a bijection?
A: In addition to the answer of spaceisdarkgreen, I want to give a full solution to the remaining c) and d).
c):

We want to show that $S_n^{<,+}\cong S_m^{<,+}$ implies $n=m$.

So let $f:\mathbb{Z}\to\mathbb{Z}$ be an $L$-isomorphism.
Then $f$ has the properties:

(I) $f(n)=m$
(II) $a<b\Leftrightarrow f(a)<f(b)$
(III) $f(a)+f(b)=f(a)+f(b)$

Suppose $n\neq m$. Without loss of generality we assume $n<m$.
If $n=0$ we have $f(0)=f(0+0)\stackrel{(III)}{=}f(0)+f(0)\Leftrightarrow 0=f(0)$
With (I) $m=0$. So $n\neq 0$.
Then we have $f(n)=\begin{cases} nf(1), n>0\\ nf(-1), n<0\end{cases}$
Where of course $nf(1)=\underbrace{f(1)+\dotso + f(1)}_{\text{n-times}}$
If $n>0$ then $m=f(n)=nf(1)$. So $n\mid m$ and $f(1)=\frac{m}{n}$.
It is $0<1<2\stackrel{(II)}{\Leftrightarrow} f(0)<f(1)<f(2)$
So $0<\frac{m}{n}<f(2)$. Now suppose that $f(1)>1$. Since $f$ is bijective there is some $k\in\mathbb{Z}$ with $f(k)=1$.
Either $0<1<k$ or $k<0<1$.
In the first case we have $0<1<k\Leftrightarrow 0<\frac{m}{n}<1$. So $m<n$. Which is a contradiction.
If $k<0<1$ then $f(k)<f(0)$ but $f(k)=1<f(0)=0$ is also a contradcition.
The case $n<0$ is worked out analogously.
So indeed $n=m$.
d):

When holds $S_n^+\cong S_m^+$

The isomorphism obvioulsy holds for $m\in\{\pm n\}$ and there are no other cases.
Like above we have $f(0)=0$ so $n\neq 0$, and $f(1)=\frac{m}{n}$. So $n\mid m$ and $n<m$.
Now we show that $m|n$:
Since $f$ is a bijection there is some $k\in\mathbb{Z}$ with $f(k)=1$.
Then $mf(k)=m\Leftrightarrow f(mk)=m\Leftrightarrow mk=n$. So $m\mid n$.
Now we have $m\mid n$ and $n\mid m$. This only holds if $\pm m=n$.
