Solve $x^x\equiv1 \pmod {14}$ I know that the order $x$ must be relatively prime to $14$ in order to have a solution so do I just check $1,3,5,9,11,13$ and see which ones raised to themselves are congruent to $1$ mod $14$? 

This answer uses results mentioned in the comments above (specifically @RobertIsrael)
You are correct that you need only consider bases which are relatively prime to $14$.  In other words, you only need to consider the classes of
$$
1,3,5,9,11,13.
$$
Next, we should compute the orders of each of these.  Using $o(n)$ for the order of $n$ modulo $14$, we have
\begin{align*}
o(1)&=1&o(3)&=6\\
o(5)&=6&o(9)&=3\\
o(11)&=3&o(13)&=2.
\end{align*}
Now, what solves $x^x\equiv 1\pmod {14}$.  Let's consider each case:


*

*If $x$ is equivalent to $1\pmod{14}$, then, since the order of $1$ is $1$, any power will do, so $\{1+14n:n\in\mathbb{Z}\}$ all satisfy $x^x\equiv 1\pmod{14}$.

*If $x$ is equivalent to $3\pmod{14}$, then, since the order of $3$ is $6$, to get $1$, you need the power to be a multiple of $6$.  In other words, $x=3+14n$ and $3+14n$ is a multiple of $6$.  This is impossible because $6$ and $14$ are even while $3$ is odd.

*If $x$ is equivalent to $5\pmod{14}$, then, since the order of $5$ is $6$, to get $1$, you need the power to be a multiple of $6$.  In other words, $x=5+14n$ and $5+14n$ is a multiple of $6$.  This is impossible because $6$ and $14$ are even while $5$ is odd.

*If $x$ is equivalent to $9\pmod{14}$, then, since the order of $9$ is $3$, to get $1$, you need the power to be a multiple of $3$.  In other words, $x=9+14n$ and $9+14n$ is a multiple of $3$ since $9$ is already a multiple of $3$.  This only happens when $n$ is a multiple of $3$, so $\{9+14\cdot 3n:n\in\mathbb{Z}\}$ are the only $x$'s equivalent to $9$ which satisfy the condition.

*If $x$ is equivalent to $11\pmod{14}$, then, since the order of $11$ is $3$, to get $1$, you need the power to be a multiple of $3$.  In other words, $x=11+14n$ and $11+14n$ is a multiple of $3$.  This only happens when $n$ is equivalent to $2\pmod 3$ because $11$ and $14$ are equivalent to $-1\pmod3$.  So, $\{11+14\cdot (2+3n):n\in\mathbb{Z}\}$ are the only $x$'s equivalent to $11$ which satisfy the condition.

*Finally, if $x$ is equivalent to $13\pmod{14}$, since the order of $13$ is $2$, to get $1$, you need the power to be a multiple of $2$.  In other words, $x=13+14n$ and $13+14n$ is a multiple of $2$.  This is impossible because $2$ and $14$ are even while $13$ is odd.
