Determine whether these algebraic structures are isomorphic 
Show that $\phi$ is an isomorphism of $\langle \mathbb{Z}, + \rangle$ with $\langle\mathbb{Z}, +\rangle$, where $\phi(n) = -n ~$for$ ~n \in \mathbb{Z}.$

I know that to show that these two binary structures are isomorphisms, I have to show that $\phi(n) = -n$ is one-to-one and onto.  I have experience finding isomorphisms of matrices in linear algebra, where I used dependent/independent matrix rows to determine whether the matrices were invertible and thus isomorphic. 
For this problem, I am unsure how to just get started, mainly showing how these binary structures are one-to-one and onto. 

exercise: Let $\langle G, + \rangle$ be any 'binary structure' (group?). Show that $\phi(g) = -g$ and $\varphi(g) = g$ are both always automorphisms on $\langle G, + \rangle$.
Let $g, g_1, g_2 \in G$ and $\phi(g) = -g$. This is kind of like an inverse property for G.
- $\phi(g)$ is onto because we can take the inverse, $\phi(-g) = g$ and $g \in G$.
- $\phi(g)$ is one-to-one because $\phi(g_1) = \phi(g_2) \rightarrow g_1 = g_2$. The $\phi$ function produces the same result for $g_1 \textrm{ and } g_2 \in G$.
- $\phi(g_1 + g_2) = -(g_1 + g_2) = (-g_1) + (-g_2) = \phi(g_1) + \phi(g_2)$ Binary structures are homomorphic(?) to each other  
Let $g, g_1, g_2 \in G$ and $\varphi(g) = g$. This is essentially the identity property for G.
- $\phi(g) = g \textrm{ and } \phi(g) = g$ (switch around the input/output but notation isn't clear) for onto check.
- $\phi(g_1) = \phi(g_2) \rightarrow g_1 = g_2$ (passes one-to-one check)
- $\phi(g_1 + g_2) = (g_1 + g_2) = (g_1) + (g_2) = \phi(g_1) + \phi(g_2)$ (homomorphism check)
Comments - That felt straight forward, but it felt more like I was adapting the mechanics of what you had done rather than strengthening the conceptual aspect
 A: I think you are viewing $\langle \mathbb{Z}, + \rangle$ as an additive group, and I think your question is to check that $\phi(n) = -n$ yields an isomorphism from $\langle \mathbb{Z}, + \rangle$ to $\langle \mathbb{Z}, + \rangle$. If not, let me know and I will edit my answer. If so, indeed we need to check that $\phi$ is one-to-one and onto, as you mentioned, and also need to check that it is a homomorphism which would mean that $\phi(n_1 + n_2) = \phi(n_1) + \phi(n_2)$. 
one-to-one:
Suppose $\phi(n_1) = \phi(n_2)$, then $-n_1 = -n_2$ which implies $n_1 = n_2$, thus by definition $\phi$ is one-to-one. 
onto:
Let $n$ be any element of $\mathbb{Z}$. See that $-n$ is also an element of $\mathbb{Z}$ and $\phi(-n) = -(-n) = n$ so that $\phi$ is onto. 
homomorphism: $\phi(n_1 + n_2) = -(n_1 + n_2) = (-n_1) + (-n_2) = \phi(n_1) + \phi(n_2)$.
So $\phi$ is indeed an isomorphism. 

remark:
When an isomorphism starts and ends on the same structure, like $\phi \colon A \to A$, we usually call $\phi$ an automorphism. Automorphism is just a special type of isomorphism.

exercise: Let $\langle G, + \rangle$ be any 'binary structure' (group?). Show that $\phi(g) = -g$ and $\varphi(g) = g$ are both always automorphisms on $\langle G, + \rangle$.
