1. Find all triples $(f, g, h)$ of injective functions from the set of real numbers to itself satisfying
$$
\begin{aligned}
f(x+f(y)) & =g(x)+h(y) \\
g(x+g(y)) & =h(x)+f(y) \\
h(x+h(y)) & =f(x)+g(y)
\end{aligned}
$$
for all real numbers $x$ and $y$. (We say a function $F$ is injective if $F(a) \neq F(b)$ for any distinct real numbers $a$ and $b$.)
Proposed by Evan Chen.
Answer. For all real numbers $x, f(x)=g(x)=h(x)=x+C$, where $C$ is an arbitrary real number.
Solution 1. Let $a, b, c$ denote the values $f(0), g(0)$ and $h(0)$. Notice that by putting $y=0$, we can get that $f(x+a)=g(x)+c$, etc. In particular, we can write
$$
h(y)=f(y-c)+b
$$
and
$$
g(x)=h(x-b)+a=f(x-b-c)+a+b
$$

So the first equation can be rewritten as
$$
f(x+f(y))=f(x-b-c)+f(y-c)+a+2 b .
$$

At this point, we may set $x=y-c-f(y)$ and cancel the resulting equal terms to obtain
$$
f(y-f(y)-(b+2 c))=-(a+2 b) .
$$

Since $f$ is injective, this implies that $y-f(y)-(b+2 c)$ is constant, so that $y-f(y)$ is constant. Thus, $f$ is linear, and $f(y)=y+a$. Similarly, $g(x)=x+b$ and $h(x)=x+c$.
Finally, we just need to notice that upon placing $x=y=0$ in all the equations, we get $2 a=b+c, 2 b=c+a$ and $2 c=a+b$, whence $a=b=c$.
So, the family of solutions is $f(x)=g(x)=h(x)=x+C$, where $C$ is an arbitrary real. One can easily verify these solutions are valid.
This problem and solution were proposed by Evan Chen.
Remark. Although it may look intimidating, this is not a very hard problem. The basic idea is to view $f(0), g(0)$ and $h(0)$ as constants, and write the first equation entirely in terms of $f(x)$, much like we would attempt to eliminate variables in a standard system of equations. At this point we still had two degrees of freedom, $x$ and $y$, so it seems likely that the result would be easy to solve. Indeed, we simply select $x$ in such a way that two of the terms cancel, and the rest is working out details.
Solution 2. First note that plugging $x=f(a), y=b ; x=f(b), y=a$ into the first gives $g(f(a))+h(b)=g(f(b))+h(a) \Longrightarrow g(f(a))-h(a)=g(f(b))-h(b)$. So $g(f(x))=h(x)+a_{1}$ for a constant $a_{1}$. Similarly, $h(g(x))=f(x)+a_{2}, f(h(x))=g(x)+a_{3}$.

Now, we will show that $h(h(x))-f(x)$ and $h(h(x))-g(x)$ are both constant. For the second, just plug in $x=0$ to the third equation. For the first, let $x=a_{3}, y=k$ in the original to get $g(f(h(k)))=h\left(a_{3}\right)+f(k)$. But $g(f(h(k)))=h(h(k))+a_{1}$, so $h(h(k))-f(k)=h\left(a_{3}\right)-a_{1}$ is constant as desired.
Now $f(x)-g(x)$ is constant, and by symmetry $g(x)-h(x)$ is also constant. Now let $g(x)=$ $f(x)+p, h(x)=f(x)+q$. Then we get:
$$
\begin{aligned}
f(x+f(y)) & =f(x)+f(y)+p+q \\
f(x+f(y)+p) & =f(x)+f(y)+q-p \\
f(x+f(y)+q) & =f(x)+f(y)+p-q
\end{aligned}
$$

Now plugging in $(x, y)$ and $(y, x)$ into the first one gives $f(x+f(y))=f(y+f(x)) \Longrightarrow$ $f(x)-x=f(y)-y$ from injectivity, $f(x)=x+c$. Plugging this in gives $2 p=q, 2 q=p, p+q=0$ so $p=q=0$ and $f(x)=x+c, g(x)=x+c, h(x)=x+c$ for a constant $c$ are the only solutions.

This second solution was suggested by David Stoner.
Solution 3. By putting $(x, y)=(0, a)$ we derive that $f(f(a))=g(0)+h(a)$ for each $a$, and the analogous counterparts for $g$ and $h$. Thus we can derive from $(x, y)=(t, g(t))$ that
$$
\begin{aligned}
h(f(t)+h(g(t))) & =f(f(t))+g(g(t)) \\
& =g(0)+h(t)+h(0)+f(t) \\
& =f(f(0))+g(t+g(t)) \\
& =h(f(0)+h(t+g(t)))
\end{aligned}
$$
holds for all $t$. Thus by injectivity of $h$ we derive that
$$
f(x)+h(g(x))=f(0)+h(x+g(x))
$$
holds for every $x$.
Now observe that placing $(x, y)=(g(a), a)$ gives
$$
g(2 g(a))=g(g(a)+g(a))=h(g(a))+f(a)
$$
while placing $(x, y)=(g(a)+a, 0)$ gives
$$
g(g(a)+a+g(0))=h(a+g(a))+f(0) .
$$

Equating this via $(*)$ and applying injectivity of $g$ again, we find that
$$
2 g(a)=g(a)+a+g(0)
$$
for each $a$, whence $g(x)=x+b$ for some real number $b$. We can now proceed as in the earlier solutions.
This third solution was suggested by Mehtaab Sawhney.
Solution 4. In the first given, let $x=a+g(0)$ and $y=b$ to obtain
$$
f(a+g(0)+f(b))=g(a+g(0))+h(b)=h(a)+h(b)+f(0) .
$$

Swapping the roles of $a$ and $b$, we discover that
$$
f(b+g(0)+f(a))=f(a+g(0)+f(b)) .
$$

But $f$ is injective; this implies $f(x)-x$ is constant, and we can the proceed as in the previous solutions.
This fourth solution was suggested by alibez.