Are these ODEs equivalent? I have the following set of ordinary differential equations:
\begin{equation}
\left\{
\begin{array}{l}
\dot{a} = f_1(a, b, c, d) \\
\dot{b} = f_2(a, b, c, d) \\
\dot{c} = f_1(c, d, a, b) \\
\dot{d} = f_2(c, d, a, b)
\end{array} 
\right.
\end{equation}
where $f_1$ and $f_2$ are two functions from $\mathbb{R}^4$ to $\mathbb{R}$.
Also, I have the following initial conditions:
\begin{equation}
\left\{
\begin{array}{l}
a(0) = \alpha \\
b(0) = \beta \\
c(0) = \alpha \\
d(0) = \beta
\end{array} 
\right.
\end{equation}
and hence, under some suitable conditions, I have the unique solutions $a(t)$, $b(t)$, $c(t)$ and $d(t)$.
Then, I pose that:
$$x(t) = \frac{a(t) + c(t)}{2} ~ \text{and} ~ y(t) = \frac{b(t) + d(t)}{2} $$
Can I prove that there exists two function, $g_1$ and $g_2$, from $\mathbb{R}^2$ to $\mathbb{R}$, such that the solution of the Cauchy problem
\begin{equation}
\left\{
\begin{array}{l}
\dot{x} = g_1(x, y) \\
x(0) = \alpha \\
\dot{y} = g_2(x, y)\\
y(0) = \beta
\end{array} 
\right.
\end{equation}
has the properties that $x(t) = a(t) = c(t)$ and $y(t) = b(t) = d(t)$?

If you can express $f_1(a,b,c,d) + f_1(c, d, a, b)$ and $f_2(a, b, c, d) + f_2(c, d, a, b)$ as a function of $(a+c)$ and $(b+d)$, then yes. If you can't, then no.
Of course if $f_1(a, b, c, d)$ and $f_2(a, b, c, d)$ are both function of $(a+c)$ and $(b+d)$ then this condition will be satisfied.
Edit: corrected by @the_candyman in a comment.
