RSA public key can decrypt the ciphertext it encrypted? For this example of my problem I used $2$ and $7$ for $p$ and $q$, and $2$ for my plaintext... but I have tried this with many different numbers and they all have the same problem.  Phi of $n$ was $6$.  My public key was $m^5~mod~14$ and my private key was $m^{11}~mod~14$.  This all worked fine.  If I encrypted my plaintext $2$ with my public key, $2^5~mod~14$, I got $4$.  If I took my ciphertext, $4$, and sent it through the same equation again, $4^5~mod~14$, I get back $2$. How can it be that my RSA public key can decrypt the ciphertext it encrypted?
The prime 2 is special, and too small.

You happened to choose an $e$ which is equal to $d,$ as I explain below.
For a proper choice of large primes, it is extremely unlikely that $d=e,$ but if it is, one can simply choose another $e$ and compute the corresponding $d.$

Also, a correction, you public key is $(n,e)=(14,5)$. It is your ciphertext which is $c=m^5~mod~14.$ For this tiny example it happens that $e=5$ satisfies $$e^2=1~(mod~\phi(14))$$ so $d=e$, thus your public and private exponents are the same, not a good thing.

Also you say your private key is $m^{11}~mod~14$, which I interpret to mean your private key $(d,p,q)$ has $d=11.$ This is a valid $d$ since $5\times 11=55=1~mod~\phi(14).$ The correct terminology is $$m=c^d~(mod ~n)$$since the letter $m$ is reserved for the message.