Let $s,t:X\to Y$ be maps in some category. Prove that $s=t$ iff the equalizer of $s$ and $t$ exists and is an isomorphism, iff the coequalizer of $s$ and $t$ exists and is an isomorphism.
Is the following proof correct?
Suppose $(E,f)$ is the equalizer of $s$ and $t$, and suppose that $f$ is an iso with inverse $F$. Then we have $sf=tf$, and composing with $F$ gives $sfF=tfF$, i.e., $s=t$. Conversely, suppose $s=t$. Let $E=X,f=id_X$. Then $sf=tf$, so $(E,f)$ is a fork. If $(E',g)$ is another fork, then define $\bar g:E'\to E$ by $\bar g=g$. Then $f\bar g=g$, and this $\bar g$ is the unique map with such property (because the property reads $id_X \bar g=g$ or $\bar g=g$). Thus $(E=X,f=id_X)$ is the coequalizer of $s$ and $t$.
For the second part, let's change the notation a bit: let $s,t:Y\to X$. Suppose $(C,f)$ is the coequalizer of $s$ and $t$ and that $f$ is an iso with inverse $F$. Since $fs=ft$, we have $Ffs=Fft$ or equivalently $s=t$. Conversely, if $s=t$, then define $C=X,f=id_X$. Claim: $(C,f)$ is the coequalizer of $s,t$. Indeed, $fs=ft$ (so $(C,f)$ is a cofork, if this is the right term), and if $(C',g:X\to C')$ is another cofork, then define $\bar g:C\to C'$ by $\bar g=g$. As in the previous case, $\bar g$ is the unique arrow that satisfies $g=\bar g f$.
Yes, your proof is correct (except for a typo at the end of the first paragraph: $(E = X, f = \mathit{id}_X)$ is an equalizer, not a coequalizer).
For the second part, I suggest that you exploit the so-called duality principle: if a statement is true in a category $\mathcal C$, its dual is true in the opposite category $\mathcal C^{\mathrm{op}}$. In this way you don't need to repeat the same argument as in the first part.