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I've been having the most difficult time with what ought to be the easiest of properties. The setup is as follows:

let $G$ be a group and $X$ some set on which $G$ acts on the left. Explicitly if $g,h\in G$ and $x \in X$ then $(gh,x) = g \cdot h \cdot x = g \cdot (h \cdot x)$ so that when $gh$ acts on $x$ we first act with $h$ and then with $g$. In the group $\mathrm{Sym} (X)$, we compose left to right so that if $\sigma, \tau \in \mathrm{Sym} (X)$ and $x \in X$ then $(\sigma \tau)(x) = \sigma (\tau (x))$ i.e., first $\tau$, then $\sigma$. I'm trying to prove that a map $\Phi : G \to \mathrm{Sym} (X)$ which I've defined is a group homomorphism. I've shown that for $g,h \in G$ and $x \in X$ we have $$ \Phi (gh)(x) = (\Phi (g) \circ \Phi (h))(x) $$ While this looks perfectly correct, that which is causing me stress is the fact that the particular group $G$ I'm working with is a fundamental group and as such there is a "handedness" to the group law: an element $g \in G$ is a homotopy class of a loop and if $h\in H$ then $gh$ means "first follow $g$, then follow $h$" (and then mod out by homotopy equivalence.) In this case the displayed equation doesn't seem quite right since on the LHS we follow $g$ first, then $h$ whereas on the RHS we follow (the image of) $h$ and then (the image of) $g$.
Left vs. right actions have really been messing with me recently!

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    What is your question, exactly? – 2017-01-17
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    Given that which I've written, is the displayed equation an example of a homomorphism or an anti-homomorphism? – 2017-01-17
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    Note that a set $X$ with a *right* $G$-action is the same as an antihomomorphism $G \to \operatorname{Sym}(X)$. – 2017-01-17

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Such a map $\Phi$ is indeed a homomorphism. You have shown that $\Phi(gh)=\Phi(g)\circ\Phi(h)$. Since $\circ$ is the group operation on $\operatorname{Sym}(X)$ ($\sigma\circ\tau$ is defined by the formula $(\sigma\circ\tau)(x)=\sigma(\tau(x))$, so it agrees with your definition of $\sigma\tau$), this says $\Phi$ preserves the group operation.