Let $C_1\!: a_1x^2 + b_1y^2 + c_1z^2 + d_1xy + e_1xz + f_1yz$ and $C_2\!: a_2x^2 + b_2y^2 + c_2z^2 + d_2xy + e_2xz + f_2yz$ be nondegenerate projective plane conics over a nonperfect field $k$ of even characteristics.
Questions:
If $C_1$ is isomorphic over $k$ to $C_2$ in algebraic geometry sense, then $C_1$ is projectively equivalent over $k$ to $C_2$?
How can I find coefficients of the projective equivalence except the brute-force search.
Does "non-degenerate" for you mean "smooth"? I will assume so. The answer to (1) is then affirmative, whereas (2) doesn't seem to have a good answer without being told how you "know" the given conics are isomorphic. The quick answer to (1) is to argue by recognizing the canonical nature of the embedding into the plane (up to projective equivalence) as being defined by the global sections of the anti-canonical bundle. But let's see that this affirmative answer is not limited to 1-dimensional smooth quadrics.
More generally, consider quadratic spaces $(V,q)$ and $(V',q)$ of dimension $n>1$ over any field $k$ and assume the associated projective quadrics $(q=0)$ and $(q'=0)$ are smooth. Over $k_s$ the quadratic spaces are conformal. Indeed, if $n$ is even then over $k_s$ there is only one possibility even up to isometry. Suppose instead that $n=2m+1$ is odd (such as $n=3$), so the non-degenerate quadratic forms over $k_s$ are given by $ax_0^2 + x_1 x_2 + \dots + x_{2m-1}x_{2m}$ for $a \in k_s^{\times}$. The class of $a$ in $k_s^{\times}/(k_s^{\times})^2$ influences the isometry class over $k_s$ when $k$ is imperfect of characteristic 2 (the restriction of the quadratic form to its defect line is $ax^2$) but in general it is not relevant to the conformal class since we can scale the quadratic form by $1/a$ and then rename $(1/a)x_{2j}$ as $x_{2j}$ to get rid of $a$.
We claim that the map of Isom-schemes $${\mathbf{Isom}}((V,q),(V',q'))/\mathbf{G}_m \rightarrow {\mathbf{Isom}}((q=0),(q'=0))$$ is an isomorphism (so passing to rational points then gives the result, by Hilbert 90). For this isomorphism assertion involving $k$-schemes, the quadratic spaces only matter up to their conformal isometry class. By Galois descent we can assume $k=k_s$, and so by moving within the unique conformal class we can further assume (for this purpose!) that $(V',q')=(V,q)$; i.e., we claim the map of $k$-group schemes $$f:{\rm{PGO}}(q)\rightarrow {\mathbf{Aut}}_{(q=0)/k}$$ is an isomorphism when $n \ne 4$. We can even assume $k = \overline{k}$ if it makes us feel better.
By design, ${\rm{PGO}}(q)$ is the automorphism scheme of the pair $(\mathbf{P}(V^{\ast}), (q=0))$ consisting of the ambient projective space $P := \mathbf{P}(V^{\ast})$ and its specified smooth quadric $D := (q=0)$. Hence, $f$ is identified with the "restriction" map $$\rho:{\mathbf{Aut}}_{(P,D)/k} \rightarrow {\mathbf{Aut}}_{D/k}$$ of Aut-schemes and we claim this map is an isomorphism when $n \ne 4$. The case $n=2$ is clear by hand (just says that the Aut-scheme of $\mathbf{P}^1$ preserving the unordered pair $\{0, \infty\}$ is the evident $\mathbf{Z}/(2)$), so we now assume $n \ge 3$ (and $n\ne 4$); the case of original interest is $n=3$.
Now $D$ is geometrically integral, so ${\rm{Pic}}_{D/k}$ exists by Grothendieck's methods, and it is a constant scheme since its tangent space ${\rm{H}}^1(D, \mathscr{O}_D)$ vanishes (a standard calculation for projective hypersurfaces). We claim that ${\rm{Pic}}(D) \simeq \mathbf{Z}$. For $n=3$ this is clear by hand since then $D \simeq \mathbf{P}^1$ (and it is false for $n=4$ because then the ruled surface $D \simeq (xy=zw) \subset \mathbf{P}^3$ has Picard group $\mathbf{Z} \oplus \mathbf{Z}$). For $n \ge 5$, so $\dim D \ge 3$, we even have that ${\rm{Pic}}(P) \rightarrow {\rm{Pic}}(D)$ is an isomorphism, due to the global Lefschetz theorem [SGA2, XII, Cor. 3.6]. (I am appealing to this heavy result solely so that all characteristics are treated on equal footing and to avoid needing to get into explicit computations; the concrete references all seem to be afraid of characteristic 2.)
The action of the $k$-group scheme ${\mathbf{Aut}}_{D/k}$ on the constant $k$-group scheme ${\rm{Pic}}_{D/k} \simeq \mathbf{Z}_k$ has to be trivial since it must preserve the unique ample generator. Note that we have not addressed whether ${\mathbf{Aut}}_{D/k}$ is smooth, and that is not necessary for the triviality of its action as just mentioned; i.e., to be rigorous while remaining ignorant about the smoothness of that Aut-scheme one has to use the usual base change formalism for coherent cohomology. (Of course, for the original case $n=3$ the smoothness of the Aut-scheme is clear by inspection.)
Let $\mathscr{L}$ be an ample line bundle on $D$ (such as the one arising from the given projective embedding $j:D \hookrightarrow P$ at the outset). Since $D$ is geometrically integral, the triviality of the action on the Picard scheme implies that ${\mathbf{Aut}}_{D/k}$ has a natural projective representation on $\Gamma(D, \mathscr{L})$. (Here I am using easy flat base change formalism for global sections to be rigorous about making this projective representation of the Aut-scheme.) Hence, we get a natural action of the $k$-group scheme $\mathbf{Aut}_{D/k}$ on the projective space $\mathbf{P}(\Gamma(D, \mathscr{L}))$.
The natural map $V^{\ast} = \Gamma(P, \mathscr{O}_P(1)) \rightarrow \Gamma(D, j^{\ast}(\mathscr{O}_P(1)))$ is an isomorphism since the quadric hypersurface $D$ in $P$ has degree $>1$. Thus, the given inclusion of $D$ into $P$ is the canonical map determined by the space of global sections of $j^{\ast}(\mathscr{O}_P(1))$, so the resulting representation of ${\mathbf{Aut}}_{D/k}$ on $\mathbf{P}(V^{\ast})=P$ must preserve $D$ and thereby defines a map of group schemes $$\iota:{\mathbf{Aut}}_{D/k} \rightarrow {\mathbf{Aut}}_{(P,D)/k}$$ whose composition with $\rho$ is the identity. But clearly $\ker \iota = 1$, so $\rho$ is an isomorphism (with $\iota$ as its inverse).
