Answer
$\{f|f(x)=x\}\cup\{f|f(x)=b-x,b\text{ real}\}$
Work Step by Step
We will determine the linear functions which are their own inverses.
Let $f(x)=mx+b$.
$$\begin{align*}
f(x)&=mx+b\quad&&\text{Write original function.}\\
y&=mx+b\quad&&\text{Replace }f(x)\text{ by }y\\
x&=my+b\quad&&\text{Switch }x\text{ and }y.\\
x-b&=my\quad&&\text{Subtract }b\text{ from each side.}\\
\dfrac{x-b}{m}&=y\quad&&\text{Divide each side by }m.
\end{align*}$$
The inverse function of $f$ is $f^{-1}(x)=\dfrac{x-b}{m}$.
In order to have $f=f^{-1}$ we must have:
$$\begin{cases}
m=\dfrac{1}{m}\\
b=-\dfrac{b}{m}.
\end{cases}$$
$$\begin{cases}
m^2=1\\
b(m+1)=0.
\end{cases}$$
$m=1\Rightarrow b=0$
$m=-1\Rightarrow b$ any real number
Therefore the family of linear functions which are their own inverses consists of the functions:
$$\{f|f(x)=x\}\cup\{f|f(x)=b-x,b\text{ real}\}$$.
