Can I say that $E[X | c] = E[c | X]$, when $c$ is a constant and $X$ is a random variable? $X$ is a random variable defined in a sample space
$c$ is a constant $\in R$ (real numbers).
 A: No to the question in the title. It can happen but that is a special case.

If $c$ is looked at as random variable defined on the same space as $X$ then $\mathbb E[X\mid c]$ is a random variable that can only take one value. This because by definition it is measurable wrt $\sigma(c)=\{\varnothing,\Omega\}$.
Further its expectation must equalize the expectation of $X$.
These two facts together leave only one choice: $\mathbb E[X\mid c]=\mathbb EX$.
Further we have $\mathbb E[c\mid X]=c$ because $\sigma(c)=\{\varnothing,\Omega\}\subseteq\sigma(X)$.
That together tells us that: $$\mathbb E[X\mid c]=\mathbb E[c\mid X]\iff \mathbb EX=c$$
A: You cannot condition on a constant, so I guess you mean constant random variable. A constant random variable is independent of any other random variable, so
$$
E[X|c]=E[X]
$$
So you're asking if $E[X]$ is equal to $c$s value.
A: Maybe translating to an applied example will help illustrate why the equality typically doesn't hold (see the other answers for the special cases in which it does). I will use conditioning a bit hand-waivingly here as an information set rather than as proper conditional distributions.
Take $c$ to be the constant $\pi$ and $X$ the outcome of a coin flip, heads or tails. If you knew that $c$ equals $3.14...$, would that alter your expectation $E(X)$? No, I suppose, so $$E(X)=E(X|c)=1/2$$ (assuming a fair coin).
Conversely, if you knew the outcome of the coin flip, would that alter your expectation regarding $E(c)$? No, I suppose, so $E(c)=E(c|X)=3.14...$
So my hunch is your professor wanted to illustrate that conditional expectations do not "commute".
