About direct limit of groups Let $G_i$ be sequence of groups for $i\in \mathbb N$ and Let $\phi_i$ be a monomorphism from $G_i$ to $G_{i+1}$.
Let $\Sigma$ be the direcet limits of $G_i$ under the embeddings of $\phi_i$.
Let $\varphi_i$ be another monomorphism from $G_i$ to $G_{i+1}$ s.t. 
$$\phi_i(G_i)=\varphi_i(G_i)$$
i.e their image are same. (but not element wise). 
If $\Sigma '$ is the direct limits of $G_i$ under the embeding of $\varphi_i$ then can we say that $\Sigma\cong \Sigma '$ ?
Note: I had asked this question there.
 A: Here is an example where they are not isomorphic (where the $G_i$ are countable abelian groups).
Write $C_k$ for the cyclic group of order $k$ and $C_k^{(I)}$ the group of finitely supported functions $I\to C_k$. 
Let $I,J$ be two disjoint infinite countable sets, and define $G=C_2^{(I)}\oplus C_4^{(J)}$, here $C_2^{(I)}$ has a basis $(e_i)_{i\in I}$ as a free $\mathbf{Z}/2\mathbf{Z}$-module and $C_4^{(J)}$ has a basis $(e_j)_{j\in J}$ and $(e_j)_{j\in K}$ as a free $\mathbf{Z}/4\mathbf{Z}$-module. Let $j_0\subset J$ be some distinguished element. Let $H$ be the subgroup of $G$ (of index 2) generated by the $e_i$ for $i\in I$, $e_j$ for $j\in J\smallsetminus\{j_0\}$, and $e_{j_0}^2$.
Let us define $G_n=G$ (I avoid to write $G_i$). We will construct two isomorphisms $\phi,\varphi:G\to H$, which are thus injective endomorphisms of $G$ with the same image, and define $\phi=\phi_n$, $\varphi=\varphi_n$ for all $n$.
Fix a bijection $c:J\to J\smallsetminus \{j_0\}$ and an element $i_0\in I$. Define $\phi(e_{i_0})=\varphi(e_{i_0})=e_{j_0}^2$, and $\phi(e_j)=\varphi(e_j)=e_{c(j)}$ for all $j\in J$. 
It remains to define $\phi$ and $\varphi$ on $C_2^{(I\smallsetminus \{i_0\})}$, using two distinct bijections $u,v:I\smallsetminus\{i_0\}\to I$ with suitable requirements. 
With no requirement, define $\phi(e_i)=e_{u(i)}$ and $\varphi(e_i)=e_{v(i)}$ for all $i\in I\smallsetminus\{i_0\}$: then $\phi$ and $\varphi$ are isomorphisms $G\to H$.
For $\varphi$ we only make the requirement that $v$ has a fixed point $i_1$ in $I\smallsetminus \{i_0\}$, this can obviously be realized.
For $\phi$ we wish that every element of $I$ is eventually mapped to $i_0$. For this, we write $I=\{i_n:n\ge 0\}$ and define $u(i_n)=i_{n-1}$ for all $n\ge 1$. Then $u$ is indeed a bijection from $I\smallsetminus \{i_0\}$ to $I$, and clearly any iterate of any point eventually lands in $\{i_0\}$.
Let now $\Sigma_\phi$ and $\Sigma_\varphi$ be the direct limits. Then $\Sigma_\phi$ has the property that every element of order 2 is a square, unlike $\Sigma_\varphi$. This is because for every element of order 2 $x$ in $G$ there exists $n$ such that $\phi^n(x)$ belongs to $C_4^{(J)}$, and hence is a square, while there exists $x$ (namely $x=e_{i_1}$) in $G$, of order 2, not a square, and fixed by $\varphi$, so $x$ is not a square in $\Sigma_\varphi$.

Edit: Here's a small variant where the direct limits are the same groups (most likely $C_4^{(\mathbf{N})}$ and $C_4^{(\mathbf{N})}\times C_2^{(I)}$ for some nonempty $I$), but where the $G_n$ are finite:
pick $G_n=C_2^n\times C_4^n$, with "basis" $e_1,\dots,e_n,f_1,\dots,f_n$. Define $H_n$ to be the subgroup of $G_{n+1}$ generated by $e_1,\dots,e_{n-1},f_{n+1}^2,f_1,\dots,f_n$. 
We define $\varphi_n(e_i)=e_i$ for $1\le i\le n-1$, $\varphi_n(e_n)=f_{n+1}^2$, $\varphi(f_i)=f_i$ for $1\le i\le n$.
We define $\phi_n(e_i)=e_{i-1}$ for $2\le i\le n$, $\varphi_n(e_1)=f_{n+1}^2$, $\varphi(f_i)=f_i$ for $1\le i\le n$.
Then it is not hard to check that $\Sigma_\phi$ is isomorphic to $C_4^{(\mathbf{N})}$ (with basis $(f_i)_{i\ge 1}$) and $\Sigma_\varphi$ is isomorphic to $ C_2^{(\mathbf{N})}\times C_4^{(\mathbf{N})}$, with "basis" $(e_i)_{i\ge 1}$ and $(f_i)_{i\ge 1}$, where $e_i$ is by definition the image of $e_i$ from $G_n$ when $n>i$.
