Direct limit isomorphism Suppose I have a directed system $(V_i, \phi_i: V_i \rightarrow V_{i+1})$, say of vector spaces $V_i$. Let $\psi_i: V_i \rightarrow V_i$ be isomorphisms. I can construct the related directed system 
$(V_i, \phi_i\circ \psi_i: V_i \rightarrow V_{i+1})$. Is it true that the direct limits $\lim_{\phi_i} V_i$ and $\lim_{\phi_i\circ \psi_i} V_i$ are isomorphic? It seems to me that you cannot directly use $\psi_i$ to get isomorphism of the directed system but maybe the limits are still isomorphic. 
Edit: As explained in Jeremy's answer below, this is false in general. What about if $\phi, \psi$ commute? ie $\phi_i \psi_i = \psi_{i+1}\phi_i$. In this case, $\psi$ induces an isomorphism of $\lim_{\phi_i} V_i$. Does this stronger assumption imply that $\lim_{\phi_i} V_i$ and $\lim_{\phi_i\circ \psi_i} V_i$ are isomorphic? In Jeremy's example, $\phi, \psi$ do not commute. 

Take $V_i=k^2$ for all $i$, with $\phi_i$ given by the matrix $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ for all $i$. Then $\varinjlim_{\phi_i}V_i$ is one dimensional.
But now take $\psi_i$ to be the isomorphism given by the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ for all $i$. Then $\varinjlim_{\phi_i\circ\psi_i}V_i$ is zero.
The answer to the supplementary question about the case where $\phi_i\psi_i=\psi_{i+1}\phi_i$ is that in this case the directed systems are isomorphic (via $\psi_i^i:V_i\to V_i$), and therefore the direct limits are isomorphic.
