Variant of what other people already said, but I find the following statement quite intuitive:

• If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$. Then all maps of stabilizers are identified with the isomorphism $A_1 \cong B_1$ in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$ induced by the rightmost square.

From this point of view it is also clear that "abelian group" is stronger than necessary. For example, the rightmost square need only be given in the category of pointed sets.

Variant of what other people already said, but I find the following statement quite intuitive:

• If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, and $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$. By exactness of the rows, all maps of stabilizers are identified with the isomorphism $\mathrm{Im}(A_1 \to A_2) \cong \mathrm{Im}(B_1\to B_2)$ induced by the leftmost square in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$ induced by the rightmost square.

From this point of view it is also clear that "abelian group" is more structure than necessary. For example, the rightmost square need only be given in the category of pointed sets.

Variant of what other people already said, but I find the following statement quite intuitive:

• If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$. Then all maps of stabilizers are identified with the isomorphism $A_1 \cong B_1$ in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$.

From this point of view it is also clear that "abelian group" is stronger than necessary. For example, the right-most square need only be given in the category of pointed sets.

Variant of what other people already said, but I find the following statement quite intuitive:

• If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$. Then all maps of stabilizers are identified with the isomorphism $A_1 \cong B_1$ in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$ induced by the rightmost square.

From this point of view it is also clear that "abelian group" is stronger than necessary. For example, the rightmost square need only be given in the category of pointed sets.