Lawvere theories: an equivalence. I'm having trouble understanding Lawvere theories (as defined below).

Definition: A Lawvere Theory is a category $\mathcal{L}$ with finite products and with a distinguished object $A$ such that every object of $\mathcal{L}$ is (isomorphic to) a finite power of $A$; that is, for any $X\in\operatorname{Ob}(\mathcal{L})$, there is an $n\in\mathbb{N}$ with $X\cong A^{n}$. The object $A$ is called the fundamental object of $\mathcal{L}$. An arrow $\omega :A^{n}\to A$ is called an $n$-ary operation (and, in particular, arrows of the type $A^{0}=1\to A$ are called constants).

Here's an exercise in Turi's Category Theory Lecture Notes:

Let $\mathbb{N}^{\text{op}}$ be the opposite category of natural numbers and all functions. Show that Lawvere theories are equivalent to product preserving functors
  $$\mathbb{N}^{\text{op}}\to\mathbf{C}$$ that are bijective on objects.

The problem: I'm not sure what to do. The $\mathbf{C}$ doesn't help as it's undefined.


My attempt: Let $\mathcal{L}:\mathbb{N}^{\text{op}}\to\mathbf{C}$ be a product preserving functor bijective on objects. Then for any $(n_{i})_{I}\in\mathbb{N}$ for any set $I$, if $\prod_{i\in I}{n_{i}}$ exits, then $\mathcal{L}\left(\prod_{i\in I}{n_{i}}\right) = \prod_{i\in I}{\mathcal{L}(n_{i})}$ and for all $m, n\in\mathbb{N}$, $c\in Ob(\mathbf{C})$,
  
  
*
  
*$\mathcal{L}(n)=\mathcal{L}(m)\Rightarrow n=m$
  
*there exists $m_{c}\in\mathbb{N}$ with $\mathcal{L}(m_{c})=c$.
  

I then set up the commutative diagram(s) for the product with an arbitrary $(f_{i})_{I}$ such that $f_{i}: k\to n_{i}$ in $\mathbb{N}^{\text{op}}$ and took $\mathcal{L}$ of everything.
Then I got stuck.
I want to force this $\mathcal{L}$ to be a Lawvere theory. Once I've done that, I'll take a Lawvere theory and try to go in the other direction.

Have I even understood the question properly? Please help. I'd prefer HINTS ONLY.
Thank you :)


Second attempt (based on the comments): The trick is to consider what happens to $1\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$. Note that the product in $\mathbb{N}^{\text{op}}$ is the coproduct in $\mathbb{N}$, so is simply addition.
Suppose $L:\mathbb{N}^{\text{op}}\to\mathbb{C}$ is a product preserving functor bijective on objects. Then for all $c\in Ob(\mathbf{C})$, there exists a unique $m_{c}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$ with $L(m_{c})=c$ and for any $m, n\in\mathbb{N}, L(m)=L(n)\Rightarrow m=n$.
Let $L(1)=A$ and note that if a product $\sum_{i\in I}{a_{i}}$ exists in $\mathbb{N}^{\text{op}}$ for $a_{i}\in Ob(\mathbb{N}^{\text{op}})=\mathbb{N}$, it is in $\mathbb{N}$ and so $I$ would be a finite set. Hence finite products must exist in $\mathbf{C}$. Now for $X\in\mathbf{C}$, $X=L(m_{X})$ for some $m_{X}\in\mathbb{N}$ so $X=L(m_{X})=L(\sum_{j=1}^{m_{X}}{1})=\prod_{j=1}^{m_{X}}{L(1)}=A^{m_{X}}$. Thus $X\cong A^{m_{X}}$. Therefore, the pair $\langle L, \mathbf{C}\rangle$ is a Lawvere theory.

Ideas for the converse: Let $L:\mathbb{N}^{\text{op}}\to\mathcal{L}$ such that $L(1)=A$ and . . .


*

*$L(u)=U\cong V=L(v)$ iff $L(u)=L(v),$

*$Y\cong A^{m}=L(m_{A^{m}})$ implies $ L(m_{A^{m}})=Y$, or

*$L(m)=Y$ if (and only if) $Y\cong A^{m}$.


I've explored these ideas and they don't seem fruitful. I would like an explicit proof now please, $\color{red}{\large\text{not just hints}}$. This is really bugging me.
 A: Let $\mathbf C$ be a Lawvere theory: i.e. a category with finite products such that there's an object $A \in \mathbf C$ for which for every other $X \in \mathbf C$ there's a $n \in \mathbb N$ such that $A^n\cong X$.
Let's consider $\mathbb N$ the category of natural numbers and all function between them.
Clearly we have a function between the objects of $\mathbb N$ and the objects of $\mathbf C$ defined as 
$$\mathcal L\colon \mathbb N \to \mathbf C,$$
with $\mathcal L(n) = A^n$ for $n \in \mathbb N \setminus\{0\}$ and $\mathcal L(0)=\bullet$ the terminal object of $\mathbf C$.
Let $f \colon n \to m$ be a morphism in $\mathbb N$, or else a function between the sets $\{0,\dots,n-1\}$ and $\{0,\dots,m-1\}$.
For every $i \in \{0,\dots,m-1\}$ we can consider the family of morphism
$\langle \pi_i^j\rangle_{j=1,\dots,n}$ where 
$\pi^i_j=1_A$ if $i \in f^{-1}(\{j\})$ other wise $\pi^i_j \colon A \to \bullet$ is the unique map in the terminal object $\bullet \in \mathbf C$.
These morphisms give us a morphism
$$\pi^i \colon A \to A^{f^{-1}(\{i\})}$$
for universal property of products
and then we get the product morphism
$$\mathcal L(f) = \prod_{i=1}^n \pi^i \colon A^n=\mathcal L(n) \to A^m=\mathcal L(m).$$
Clearly if $f=1_n$ for a $n$ then for every $i \in n$ we have $1_n^{-1}(\{i\})=\{i\}$ and so the $\pi^i=1_A$ for every $i$ and so $\mathcal L(1_n)=1_{A^n}$.
Doing the calculations you can prove also that $\mathcal L(g \circ f)=\mathcal L(f) \circ \mathcal L(g)$, for every pair $f \colon n \to m$ and $g \colon m \to k$, of course the calculation are a little complicated (I  would rather not to write here).
This functor is clearly product preserving, indeed for every $n \in \mathbb N$ we have that $\mathcal L(n)=A^n$ which is the n-fold product, and  every projection of $\mathbb N^\text{op}$ i.e. a map $p \colon 1 \to n$ in $\mathbb N$ we have that 
$$\mathcal L(p) = \prod_{i=1}^n \pi^i \colon A^n \to A,$$
where every $\pi^i \colon A \to \bullet$ for $i \not \in \text{Im }p(0)$ and $\pi^i = 1_A$ for $i \in \text{Im }p(0)$.
So $\mathcal L(p)$ is the $p(0)$-th projection of the product $A^n$, of course now one should verify that these data verify the universal property which involve again some long calculations.
Forgive the lack of some details, but I think that adding them wouldn't make the answer more clear.
Hope this helps.
